Please Login to access more options.


Today

« December 2016 »

Sun

Mon

Tue

Wed

Thu

Fri

Sat

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

31

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 $(G/N)/(M/N)\approx G/M$.
Note that for the first problem, you only need to define a map from $K$ to $KN/N$, as the $K\cap N$ part should automatically show up when you compute the kernel and use the first isomorphism theorem. What map should you try? How about sending $k$ to the coset $Nk$?
For the third isomorphism theorem, you could define a map from $G$ to $(G/N)/(M/N)$ or from $G/N$ to $G/M$, and then in either case show that the kernel of your map is the missing denominator.
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.

  1. $n=8=2^3$
  2. $n=48 =2^43 $
  3. $n=500 = 2^25^3$
  4. 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