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Problem 101 (Homomorphisms Preserve Normal Subgroups)
Suppose that $f:G\to H$ is a homomorphism. Use The Normal Subgroup Test to prove the following:
- If $N$ is normal in $G$, then $f(N)$ is normal in $f(G)$.
- If $B$ is normal in $H$, then $f^{-1}(B)$ is normal in $G$.
At this point, we've proved quite a large collection of facts about homomorphisms and what they preserve. I strongly suggest that you look at pages 202-204 of your text to see a list of these properties. Then read the remarks on page 204.
The following two facts follow immediately from using the properties of homomorphisms. The first fact shows that any time you have a subgroup that contains half the elements of the group, that subgroup must be a normal subgroup.
Problem 102 (Subgroups Of Index 2 Are Normal)
Suppose that $H$ is a subgroup of $G$ with index $|G:H|=2$. Recall that the index of $H$ in $G$ is the number of distinct cosets of $H$ in $G$.
- Build a surjective homomorphism from $G$ to $\mathbb{Z}_2$.
- Show that $H$ is normal in $G$.
This next fact shows that any time you have subgroup in a factor group $G/N$, it corresponds to a subgroup $H$ in $G$ that contains $N$.
Problem 103 (Subgroups Of A Quotient Group Correspond To Subgroups Of The Original Group)
Suppose that $N$ is a normal subgroup of $G$ and that $B$ is a subgroup of $G/N$. Prove the following:
- There exists a subgroup $H\leq G$ such that $H/N=B$.
- If we know that $n=|N|$ and $m=|B|$ (so $n$ is the order of $N$ in $G$, and $m$ is the order of $B$ in $G/N$), then $G$ must have a subgroup $H$ of order $nm$.
As a suggestion, consider the homomorphism $f:G\to G/N$ given by $f(g)=Ng$ and use some properties of homomorphisms.
To tackle the next problem, you'll want to rely on the fact that $|(g,h)|=\text{lcm}(|g|,|h|)$, together with the fact that two integers are relatively prime if and only if their product is their least common multiple.
Problem 104 (A Direct Product Of Cyclic Groups Is Cyclic If And Only If The Groups Have Relatively Prime Orders)
Suppose that $G$ and $H$ are finite cyclic groups. Prove that $G\oplus H$ is cyclic if and only if $|G|$ and $|H|$ are relatively prime.
Exercise (A Direct Product Of Cyclic Groups Is Cyclic If And Only If The Groups Have Relatively Prime Orders)
Use induction to extend the previous result to the external direct product of $n$ cyclic groups. So prove that if $G_1,G_2,\ldots,G_n$ are $n$ cyclic groups, then $G_1\oplus G_2\oplus \cdots\oplus G_n$ is cyclic if and only if $|G_i|$ and $|G_j|$ are relatively prime when $i\neq j$.
Click to see a hint.
This exercise require that you use induction on an "if $p$ then $q$" statement. In addition, the $q$ part of the statement involves an if and only if. You must be very careful how you tackle this problem.
Problem 105 (An Isomorphism From $U(st)$ To $U(s)\oplus U(t)$ When $s$ And $t$ Are Relatively Prime)
Recall that $\varphi$, the Euler-phi function is defined by $\varphi(n)=|U(n)|$, the order of $U(n)$. Suppose $n=st$ where $s$ and $t$ are relatively prime. In this problem you'll show that $\varphi(st)=\varphi(s)\varphi(t)$.
- Show that $f:U(st)\to U(s)\oplus U(t)$ defined by $f(x)=(x\mod s,x\mod t)$ is an isomorphism.
- Why does $\varphi(st)=\varphi(s)\varphi(t)$ when $s$ and $t$ are relatively prime?
- Use the results above to compute $\varphi(17\cdot 19)$ and $\varphi(15\cdot 63)$.
It's time to use the power of factor groups, together with induction, to prove the first of many powerful theorems that apply to finite groups. The key lies in the fact that the order of a factor group $G/N$ will always be less than the order of $G$, provided $N$ is not trivial. This means that if we use induction on the order of a group, then if we can use a factor group to pull back information to a larger group, induction will provide us with precisely the tool we need to prove some really powerful theorems about groups.
The next theorem is often called Cauchy's theorem (for Abelian groups). The more general version of Cauchy's theorem states that if $p$ is a prime that divides the order of the group, then $G$ must have an element of order $p$. Because it's an Abelian group, every subgroup is normal, which means we can quickly create factor groups any time we obtain a subgroup. If we remove the Abelian condition, then we have to prove that a subgroup is normal before we can create a factor group. The Abelian condition is a simplifying assumption, which can be removed with much more work.
Problem 106 (Abelian Groups Have An Element Of Order $p$ For Every Prime That Divides The Order Of The Group)
Let $G$ be an Abelian group and let $p$ be a prime. Prove that if $p$ divides the order of $G$, then there exists $a\in G$ with order $p$.
Hint: This problem is done by induction on the order of $G$.
For each integer $k\geq 2$, let $P(k)$ be the statement, "For every positive integer $n<k$, if $|G|=n$ and $p$ divides $|G|$ then $G$ has an element of order $p$." Use induction to prove that $P(k)$ is a true statement for every integer $k\geq 2$. This will require you to do two things.
- Is the statement $P(2)$ true? Is the statement $P(3)$ true?
- If you assume for some specific $k\geq 2$ that the statement $P(k)$ is true, then prove that the statement $P(k+1)$ is true. To prove $P(k+1)$ is a true statement, you have to prove that if $n<k+1$ and $|G|$ has order $n$ with $p$ dividing $G$, then $G$ has an element of order $p$. This is quite a lot of facts to prove. However, the induction assumption should prove every single one of them instantly except for when $n=k$. So you must show that if $|G|=k$ and $p$ divides $|G|$, then $G$ has an element of order $p$. You know for sure that $G$ has an element of some order other than 1, say $m=|a|$, which means that it has an element $b$ of some prime order $q$ (why?). If $q=p$ we're done. Otherwise, consider the factor group $G/\left<b\right>$. What is the order of $G/\left<b\right>$? Does $p$ divide $G/\left<b\right>$? What does the induction assumption then tell you? How does problem 103 help?
Problem 101 (Homomorphisms Preserve Normal Subgroups)
Suppose that $f:G\to H$ is a homomorphism. Use The Normal Subgroup Test to prove the following:
- If $N$ is normal in $G$, then $f(N)$ is normal in $f(G)$.
- If $B$ is normal in $H$, then $f^{-1}(B)$ is normal in $G$.
At this point, we've proved quite a large collection of facts about homomorphisms and what they preserve. I strongly suggest that you look at pages 202-204 of your text to see a list of these properties. Then read the remarks on page 204.
The following two facts follow immediately from using the properties of homomorphisms. The first fact shows that any time you have a subgroup that contains half the elements of the group, that subgroup must be a normal subgroup.
Problem 102 (Subgroups Of Index 2 Are Normal)
Suppose that $H$ is a subgroup of $G$ with index $|G:H|=2$. Recall that the index of $H$ in $G$ is the number of distinct cosets of $H$ in $G$.
- Build a surjective homomorphism from $G$ to $\mathbb{Z}_2$.
- Show that $H$ is normal in $G$.
This next fact shows that any time you have subgroup in a factor group $G/N$, it corresponds to a subgroup $H$ in $G$ that contains $N$.
Problem 103 (Subgroups Of A Quotient Group Correspond To Subgroups Of The Original Group)
Suppose that $N$ is a normal subgroup of $G$ and that $B$ is a subgroup of $G/N$. Prove the following:
- There exists a subgroup $H\leq G$ such that $H/N=B$.
- If we know that $n=|N|$ and $m=|B|$ (so $n$ is the order of $N$ in $G$, and $m$ is the order of $B$ in $G/N$), then $G$ must have a subgroup $H$ of order $nm$.
As a suggestion, consider the homomorphism $f:G\to G/N$ given by $f(g)=Ng$ and use some properties of homomorphisms.
To tackle the next problem, you'll want to rely on the fact that $|(g,h)|=\text{lcm}(|g|,|h|)$, together with the fact that two integers are relatively prime if and only if their product is their least common multiple.
Problem 104 (A Direct Product Of Cyclic Groups Is Cyclic If And Only If The Groups Have Relatively Prime Orders)
Suppose that $G$ and $H$ are finite cyclic groups. Prove that $G\oplus H$ is cyclic if and only if $|G|$ and $|H|$ are relatively prime.
Exercise (A Direct Product Of Cyclic Groups Is Cyclic If And Only If The Groups Have Relatively Prime Orders)
Use induction to extend the previous result to the external direct product of $n$ cyclic groups. So prove that if $G_1,G_2,\ldots,G_n$ are $n$ cyclic groups, then $G_1\oplus G_2\oplus \cdots\oplus G_n$ is cyclic if and only if $|G_i|$ and $|G_j|$ are relatively prime when $i\neq j$.
Click to see a hint.
This exercise require that you use induction on an "if $p$ then $q$" statement. In addition, the $q$ part of the statement involves an if and only if. You must be very careful how you tackle this problem.
Problem 105 (An Isomorphism From $U(st)$ To $U(s)\oplus U(t)$ When $s$ And $t$ Are Relatively Prime)
Recall that $\varphi$, the Euler-phi function is defined by $\varphi(n)=|U(n)|$, the order of $U(n)$. Suppose $n=st$ where $s$ and $t$ are relatively prime. In this problem you'll show that $\varphi(st)=\varphi(s)\varphi(t)$.
- Show that $f:U(st)\to U(s)\oplus U(t)$ defined by $f(x)=(x\mod s,x\mod t)$ is an isomorphism.
- Why does $\varphi(st)=\varphi(s)\varphi(t)$ when $s$ and $t$ are relatively prime?
- Use the results above to compute $\varphi(17\cdot 19)$ and $\varphi(15\cdot 63)$.
It's time to use the power of factor groups, together with induction, to prove the first of many powerful theorems that apply to finite groups. The key lies in the fact that the order of a factor group $G/N$ will always be less than the order of $G$, provided $N$ is not trivial. This means that if we use induction on the order of a group, then if we can use a factor group to pull back information to a larger group, induction will provide us with precisely the tool we need to prove some really powerful theorems about groups.
The next theorem is often called Cauchy's theorem (for Abelian groups). The more general version of Cauchy's theorem states that if $p$ is a prime that divides the order of the group, then $G$ must have an element of order $p$. Because it's an Abelian group, every subgroup is normal, which means we can quickly create factor groups any time we obtain a subgroup. If we remove the Abelian condition, then we have to prove that a subgroup is normal before we can create a factor group. The Abelian condition is a simplifying assumption, which can be removed with much more work.
Problem 106 (Abelian Groups Have An Element Of Order $p$ For Every Prime That Divides The Order Of The Group)
Let $G$ be an Abelian group and let $p$ be a prime. Prove that if $p$ divides the order of $G$, then there exists $a\in G$ with order $p$.
Hint: This problem is done by induction on the order of $G$.
For each integer $k\geq 2$, let $P(k)$ be the statement, "For every positive integer $n<k$, if $|G|=n$ and $p$ divides $|G|$ then $G$ has an element of order $p$." Use induction to prove that $P(k)$ is a true statement for every integer $k\geq 2$. This will require you to do two things.
- Is the statement $P(2)$ true? Is the statement $P(3)$ true?
- If you assume for some specific $k\geq 2$ that the statement $P(k)$ is true, then prove that the statement $P(k+1)$ is true. To prove $P(k+1)$ is a true statement, you have to prove that if $n<k+1$ and $|G|$ has order $n$ with $p$ dividing $G$, then $G$ has an element of order $p$. This is quite a lot of facts to prove. However, the induction assumption should prove every single one of them instantly except for when $n=k$. So you must show that if $|G|=k$ and $p$ divides $|G|$, then $G$ has an element of order $p$. You know for sure that $G$ has an element of some order other than 1, say $m=|a|$, which means that it has an element $b$ of some prime order $q$ (why?). If $q=p$ we're done. Otherwise, consider the factor group $G/\left<b\right>$. What is the order of $G/\left<b\right>$? Does $p$ divide $G/\left<b\right>$? What does the induction assumption then tell you? How does problem 103 help?
Problem. Factor Group Practice
Please open your text to chapter 9 and complete problems 14, 17, 18, 24, 25, 27, 28. Do as many of these as are valuable to your continued learning. Check your work with the solutions below.
Click For Some Solutions.
The odd numbered problems have partial solutions in the back of the text.
Exercise 14
First, remember that the identity in a factor group $G/H$ is the coset $H$. So we need to know how many times we have to add the coset $14+\left<8\right>$ to itself before we obtain $\left<8\right>$. This is equivalent to asking, "how many times must we add 14 to itself before we reach a multiple of 8?" We could just notice that the multiples of 14 are 14, 28, 42, 56, and so the 4th one gets us to a multiple of 8. This means the order is 4.
We could also simplify the problem first, before computing the order. Since $14+\left<8\right>=6+8+\left<8\right>$, and because $8+\left<8\right> = \left<8\right>$ (cosets absorb their own elements) we can just write $14+\left<8\right>=6+\left<8\right>$. To find the order, we just need to ask how many times we have to add 6 to itself until we arrive at a multiple of 8. The least common multiple is 24, which means the order of the coset is 4.
Exercise 17
Let $G=Z/\left<20\right>$ and $H=\left<4\right>/\left<20\right>$. Then we have $$H = \{4+\left<20\right>, \quad 8+\left<20\right>, \quad 12+\left<20\right>, \quad 16+\left<20\right>, \quad 0+\left<20\right>\}.$$ The elements of $G/H$ are $$H, \quad (1+\left<20\right>)+H, \quad (2+\left<20\right>)+H, \quad (3+\left<20\right>)+H\}.$$
Exercise 18
There are 60 elements in $\mathbb{Z}_60$. The subgroup generated by 15 has 4 elements. So there are 60/4=15 elements in the factor group.
Exercise 24
Let's start by listing the cosets of $H = \left<(2,2\right>$ in the group $G=\mathbb{Z}_{4}\oplus \mathbb{Z}_{12}$. The cosets are $$ \begin{align} H &= \left<(2,2\right> =\{(0,0),(2,2),(0,4),(2,6),(0,8),(2,10)\} \\ (1,0)+H &= \{(1,0),(3,2),(1,4),(3,6),(1,8),(3,10)\} \\ (2,0)+H &= \{(2,0),(0,2),(2,4),(0,6),(2,8),(0,10)\} \\ (3,0)+H &= \{(3,0),(1,2),(3,4),(1,6),(3,8),(1,10)\} \\ (0.1)+H &= \{(0,1),(2,3),(0,5),(2,7),(0,9),(2,11)\} \\ (1.1)+H &= \{(1,1),(3,3),(1,5),(3,7),(1,9),(3,11)\} \\ (2.1)+H &= \{(2,1),(0,3),(2,5),(0,7),(2,9),(0,11)\} \\ (3.1)+H &= \{(3,1),(1,3),(3,5),(1,7),(3,9),(1,11)\}. \end{align} $$ If you just ignore the plus $H$ on the end of each coset, you should see that this group is basically the same as $\mathbb{Z}_4\oplus \mathbb{Z}_2$. There are clearly no elements of order 8, which means is not isomorphic to $\mathbb{Z}_8$. There are elements of order $4$, which means it's not isomorphic to $\mathbb{Z}_2\oplus \mathbb{Z}_2\oplus \mathbb{Z}_2$.
Exercise 25
If we list out the cosets of $H=\{1,31\}$ in the group $U(32) = \{1,3,5,7,9,11,13,1,17,19,21,23,25,27,29,31\}$, we obtain (noting that 31 and $-1$ are equivalent mod 32 to simply computations) $$ \begin{align} H &= \{1,31\}\\ 3H &= \{3,29\}\\ 5H &= \{5,37\}\\ 7H &= \{7,25\}\\ 9H &= \{9,23\}\\ 11H &= \{11,21\}\\ 13H &= \{13,19\}\\ 15H &= \{15,17\}. \end{align} $$ We now need to compute the order of these cosets. The only possible orders are 1, 2, 4, and 8, because the order of the factor group is 8 and each element must have an order that divides the order of the group. The powers of 3, taken mod 32, are 3, 9, 27=-5, -15=17, and so on. Notice that this is enough to rule out 1,2, and 4 as the order of $3H$. This means that $3H$ has order 8, and as such generates the entire group. This means that the group $G/H$ must be isomorphic to $\mathbb{Z}_8$.
Exercise 27
We know $H$ and $K$ are isomorphic, because they both have prime order 2, and hence are both cyclic. See the book for the reason why the factor groups are not isomorphic.
Exercise 28
Similar to the previous. The point to these last two problems is to show you that even if you know $H$ and $K$ are isomorphic, you do not know that $G/H$ and $G/K$ are isomorphic. They might be, but they don't have to be.
You'll want to use the The First Isomorphism Theorem to prove both of the following.
Theorem (The First Isomorphism Theorem)
Let $f:G\to H$ be a surjective homomorphism with kernel $K$. Because we know that $f(x)=f(y)$ for any $y\in Kx$ (elements in the same coset of the kernel have the same image under $f$), then we can define a map $\phi:G/K\to H$ by defining $\phi(Kg)=f(g)$. This map $\phi$ is always an isomorphism.
Problem. The Second and Third Isomorphism Theorems
Use The First Isomorphism Theorem to prove each of the following. In both cases, you just need to define a surjective map, compute the kernel, and then quote the theorem.
- (Second Isomorphism Theorem) Suppose that $K\leq G$ and $N\trianglelefteq G$. Prove that $K/(K\cap N)\approx KN/N$.
- (Third Isomorphism Theorem) Suppose that $M$ and $N$ are normal subgroups of $G$ with $N\leq M$. Prove that $(G/N)/(M/N)\approx G/M$.
Theorem. Fundamental Theorem of Finite Abelian Groups
Every finite Abelian group $G$ is the direct product of cyclic groups. In particular, this direct product can be expressed as a direct product of cyclic groups with prime-power order, which means $$G\approx \mathbb{Z}_{p_1^{n_1}}\oplus \mathbb{Z}_{p_2^{n_2}}\oplus\cdots \oplus \mathbb{Z}_{p_k^{n_k}} .$$ When expressed in this manner, the number of terms in the direct product and the orders of the cyclic groups is uniquely determined by the group.
Problem.Isomorphism Classes of Abelian Groups of various orders
Start by reading pages 212-215 in your text. In each situation below, you are given an integer $n$. State all the different isomorphism classes of Abelian groups of the given order.
- $n=8=2^3$
- $n=48 =2^43 $
- $n=500 = 2^25^3$
- How many (don't list them all) isomorphism classes are there for a group of order $n=2^2\cdot 3^3\cdot 5\cdot 7^5\cdot 11^4$?
When $n=8=2^3$, the isomorphism classes are $$ \mathbb{Z}_2\oplus \mathbb{Z}_2\oplus\mathbb{Z}_2, \quad \mathbb{Z}_4\oplus \mathbb{Z}_2, \quad \text{ and }\quad \mathbb{Z}_8. $$
Problem.Practice with The Fundamental Theorem
Please complete problems 1-10 in chapter 11 of your text.
Click for some solutions.
Problem 1
The smallest such integer is $n=4$. The two groups are $\mathbb{Z}_4$ and $\mathbb{Z}_2\oplus\mathbb{Z}_2$.
Problem 2
The smallest such integer is $n=8$. The three groups are $\mathbb{Z}_8$, $\mathbb{Z}_4\oplus\mathbb{Z}_2$, and $\mathbb{Z}_3\oplus\mathbb{Z}_2\oplus\mathbb{Z}_2$.
Problem 3
The smallest such integer is $n=2^2\cdot 3^2 = 36$. The four groups are $\mathbb{Z}_4\oplus \mathbb{Z}_9$, $\mathbb{Z}_4\oplus\mathbb{Z}_3\oplus \mathbb{Z}_3$, $\mathbb{Z}_2\oplus\mathbb{Z}_2\oplus \mathbb{Z}_9$, and $\mathbb{Z}_2\oplus\mathbb{Z}_2\oplus\mathbb{Z}_3\oplus \mathbb{Z}_3$. You have to pick different primes to obtain this one.
Problem 4
In $\mathbb{Z}_{16}$ there is only one element of order 2, namely the integer 8. There are 2 elements of order 4, namely the integers 4 and 12.
In $\mathbb{Z}_{8}\oplus \mathbb{Z}_2$ we can obtain an element of order 2 by picking an element $(x,y)$ such that either $x$ or $y$ has order 2 and the other has order 1 or 2. There is one element in each group that has order 2. The elements of order 2 are $(4,0)$, $(4,1)$, and $(0,1)$. There are three such elements. To compute the elements of order 4, we must pick an element of the form $(x,y)$ where either $x$ or $y$ has order $4$, and the other has order 1, 2, or 4. The elements of order 4 are hence $(2,0),(6,0),(2,1),(6,1)$. That's it.
In $\mathbb{Z}_{4}\oplus \mathbb{Z}_4$ we can obtain an element of order 2 by picking an element $(x,y)$ such that either $x$ or $y$ has order 2 and the other has order 1 or 2. There is one element in each group that has order 2. The elements of order 2 are $(2,0)$, $(2,2)$, and $(0,2)$. There are three such elements. To compute the elements of order 4, we must pick an element of the form $(x,y)$ where either $x$ or $y$ has order $4$, and the other has order 1, 2, or 4. The elements of order 4 where we require the first component to have order 4 are $(1,0),(3,0),(1,1),(3,1),(1,2),(3,2),(1,3),(3,3),$. We also must count the elements where the first component does not have order 4, but the second does. This gives us the additional elements $(0,1),(0,3),(2,1),(2,3)$. This gives us 12 elements of order 4. A simpler computation could have just noticed that since there is 1 element of order 1 and 3 elements of order 2, and no element has order 8 or 16, then the remaining 12 elements must have order 4.
In $\mathbb{Z}_{4}\oplus \mathbb{Z}_2\oplus \mathbb{Z}_2$ we can obtain an element of order 2 by picking an element $(x,y,z)$ such that either $x$, $y$, or $z$ has order 2 and the others have orders 1 or 2. There is one element in each group that has order 2. There are 4 elements of the form $(2,x,y)$ that have order 2 (as there are two choices for $y$ and $z$). There are two choices of the form $(0,2,z)$, and one choice of the form $(0,0,2)$. This gives a total of 7 elements of order 2. To obtain an element of order 4, we must choose $x=1,3$ and then allow $y$ or $z$ to be anything. This gives $8=2\cdot 2\cdot 2$ elements of order 4. Again, we could have noticed that with 1 element of order 1 and 7 elements of order 2, there must be 8 elements of order 4.
Problem 5
Since $45=5\cdot 9$, there are two isomorphism classes of Abelian groups of order 45. They are $\mathbb{Z}_9\oplus \mathbb{Z}_5\approx \mathbb{Z}_{45}$ and $\mathbb{Z}_3\oplus\mathbb{Z}_3\oplus \mathbb{Z}_5\approx \mathbb{Z}_{15}\oplus \mathbb{Z}_3$. Remember that we can combine cyclic groups when the orders are relatively prime and obtain a cyclic group. Notice that in each case, we can obtain an element of order 15, namely $3\ in \mathbb{Z}_{45}$ and $(1,0)\in \mathbb{Z}_{15}\oplus \mathbb{Z}_3$.
Problem 6
Notice that $n=108=54\cdot 2 = 27\cdot 2^2 = 2^2\cdot 3^3$. There are two isomorphism classes for groups of order $2^2$, and three isomorphism classes for groups of order $3^3$. So there are a total of 6 isomorphism classes of groups of order 108. The cyclic group $\mathbb{Z}_{108}\approx \mathbb{Z}_{4}\oplus \mathbb{Z}_{27}$ has exactly one subgroup of order $3$. Since we are after subgroups of order 3, we can do whatever we want to the part related to the prime number 2, and it won't change the answer, as any element that includes something nonzero from any factors not divisible by 2 will force the element to have even order (recall that the order of an element is the least common multiple of the orders or the elements in each factor). This means the group $\mathbb{Z}_2\oplus \mathbb{Z}_2\oplus \mathbb{Z}_{27}$ has a single subgroup of order 3.
Problem 7
Continuing from problem 6, we notice that $\mathbb{Z}_{4}\oplus \mathbb{Z}_{9}\oplus \mathbb{Z}_{3}$ has several subgroups of order 3. Because any any subgroup of order 3 is cyclic, all we have to do is count the number of elements of order 3. So let $(x,y,z)$ be an element of $\mathbb{Z}_{4}\oplus \mathbb{Z}_{9}\oplus \mathbb{Z}_{3}$. Then $x=0$, otherwise the order is divisible by 2. The orders of $y$ and $z$ must be 1 or 3, with at least 1 of them being 3. So if $x=3$ or $x=6$, then $y\in\{0,1,2\}$. This gives 6 elements of order 3. If $x=0$, then $y\in \{1,2\}$. which gives another 2 elements of order 3. There are a total of 8 elements of order 3. Each subgroup of order 3 contains 2 of these elements, and as such there are 4 subgroups of order 3. A similar computation for $\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}\oplus \mathbb{Z}_{9}\oplus \mathbb{Z}_{3}$ gives the same result.
Problem 8
We now examine the groups $\mathbb{Z}_{4}\oplus \mathbb{Z}_{3}\oplus\mathbb{Z}_{3}\oplus \mathbb{Z}_{3}$ and $\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}\oplus \mathbb{Z}_{3}\oplus\mathbb{Z}_{3}\oplus \mathbb{Z}_{3}$. We just need to count the number of elements of order in $\mathbb{Z}_{3}\oplus\mathbb{Z}_{3}\oplus \mathbb{Z}_{3}$. This is simple, as every element other than the identity must have order 3, so there are 26 elements of order 3. This means there are 13 subgroups of order 3 in either group.
Problem 9
We know that $120=2^3\cdot 3\cdot 5$. This means that $G$ is isomorphic to one of the following groups: $$\mathbb{Z}_8\oplus\mathbb{Z}_3\oplus\mathbb{Z}_5, \mathbb{Z}_4\oplus\mathbb{Z}_2\oplus\mathbb{Z}_3\oplus\mathbb{Z}_5, \mathbb{Z}_2\oplus\mathbb{Z}_2\oplus\mathbb{Z}_2\oplus\mathbb{Z}_3\oplus\mathbb{Z}_5.$$ The first group has one element of order 2. The last group has 7 elements of order 2. This leaves the middle group as the only possible answer.
Problem 10
We first compute $36\cdot 10 = 2^2\cdot 3^2\cdot 2\cdot 5=2^3\cdot 3^2\cdot 5$. There are 6 possible isomorphism classes. See page 220 in your text for the possible answers, where they list the possible isomorphism classes of groups of order $2^3\cdot 7^2\cdot 3$.
For more problems, see AllProblems