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The next problem has you look at a very specific homomorphism from $G\times H$ to $G$. It's the start of a key idea that shows $(G\times H)/H\approx G$, namely that if you first create an external product, and then decide later to annihilate one of your factors, then you will obtain the other factor. It's basically an extension of the division fact from integers that says $(xy)/y=x$. A similar result shows that $(G\times H)/G\approx H$.

Problem (Collapsing A Factor Of An External Direct Product Yields The Other Factor, or $(G\times H)/H\approx G$)

Let $G$ and $H$ be groups, and consider the projection map $\pi_1:G\times H\to G$ defined by $\pi_1(g,h)=g$. We call this the projection map $\pi_1$ because it takes the element $(g,h)$ and returns the first component. The projection map $\pi_2$ would return the second component $h$.

  1. Prove that this projection map is a surjective homomorphism.
  2. What is the kernel of $\pi_1$, in other words what is $f^{-1}(e_G)$?
  3. Why is $(G\times H)/(\{e_G\}\times H)\approx G$?
  4. Show that $i_H:\{e_G\}\times H\to H$ defined by $i_H(e_G,h)=h$ is an isomorphism.
Because of the work above, we'll often write $(G\times H)/H\approx G$, even though $H$ is not technically a subgroup of $G\times H$, rather $\{e_g\}\times H$ is a subgroup of $H$. However, it's ugly to write $(G\times H)/(\{e_G\}\times H)\approx G$ and much prettier to just write $(G\times H)/H\approx G$, which we justify doing because $H\approx \{e_G\}\times H$.

The next problem has you practice working with orders of elements in external direct products, and recognizing when two groups are, or are not, isomorphic. Much of what we'll be doing in the next two weeks has to do with determining when two groups are, or are not, isomorphic.

Problem (Which Groups Of Order 60 Are Isomorphic)

Prove or disprove each of the following. Either build an isomorphism, or show that no such isomorphism exists.

  1. $\mathbb{Z}_4\oplus Z_{15}\approx \mathbb{Z}_{6}\oplus \mathbb{Z}_{10}$
  2. $\mathbb{Z}_4\oplus \mathbb{Z}_{15}\approx \mathbb{Z}_{20}\oplus \mathbb{Z}_{3}$
  3. $D_{20}\oplus \mathbb{Z}_{3}\approx D_{60}$ (Recall that $D_{2n}$ is the automorphisms of a regular $n$-gon.)
  4. $D_{20}\oplus \mathbb{Z}_{3}\approx \mathbb{Z}_{12}\oplus \mathbb{Z}_5$

Click to see a hint.

Consider orders of elements. What's the largest possible order in each group? How many elements of order 2 does each group have? If you find a mismatch between groups, they cannot be isomorphic.

To prove that $H$ is a subgroup of $G$, we have been using two different options, namely The Subgroup Test and The Finite Subgroup Test. Our version of the subgroup test requires that you show (1) that $H$ is nonempty, (2) that $H$ is closed under the operation, and (3) that $H$ is closed under taking inverse. There is another test that can often result in shorter proofs, though not always easier to use. It's called the "One-Step Subgroup Test" and requires you show that $H$ is nonempty and that $ab^{-1}\in H$ whenever $a$ and $b$ are in $H$.

Problem (One Step Subgroup Test)

Suppose that $H$ is a nonempty subset of a group $G$. Suppose that $ab^{-1}\in H$ whenever $a,b\in H$. Prove that $H$ is a subgroup of $G$.

Problem (If The Factor Group Of G By The Center Is Cyclic Then G Is Abelian)

Suppose that $G$ is a group and $Z(G)$ is the center of $G$. Prove that if $G/Z(G)$ is cyclic, then $G$ is Abelian.

Definition (Automorphisms And Inner Automorphisms)

Let $G$ be a group.

  • An automorphism of $G$ is an isomorphism from $G$ to $G$.
  • We write $\text{Aut}(G)$ to represent the set of all automorphisms of $G$.
  • The function $\phi_g:G\to G$ defined by $\phi_g(x)=gxg^{-1}$ is called an inner automorphism of $G$.
  • We write $\text{Inn}(G)$ to represent the set of all inner automorphisms of $G$.

Problem ($\text{Inn}(G)$ is a normal subgroup of $\text{Aut}(G)$)

Let $G$ be a group. Prove the following:

  1. $\text{Aut}(G)$ is a group,
  2. $\text{Inn}(G)$ is a subgroup of $\text{Aut}(G)$, and
  3. $\text{Inn}(G)$ is a normal subgroup of $\text{Aut}(G)$.

Problem ($G/Z(G)$ is isomorphic to $\text{Inn}(G)$)

Suppose that $G$ is a group. Let $f:G \to \text{Inn}(G)$ be defined by $f(x)=\phi_x$.

  1. Show that $f$ is a homomorphism with kernel $Z(G)$.
  2. Then prove that $G/Z(G)$ is isomorphic to $\text{Inn}(G)$.
  3. Compute $\text{Inn}(G)$ for any Abelian group $G$ and then for $G=D_8$.
  4. Why is $D_{10}$ isomorphic to $\text{Inn}(D_{10})$?

Problem. Factor Group Practice

Please open your text to chapter 9 and complete problems 14, 15, 17, 18, 19, 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 15

Remember that $U_5(105)$ is the collection of elements $x\in U(105)$ such that $x\mod 5=1$. We need to find the order of $4U_5(105)$. Since $4^2=16$, and we know that $16\mod 5=1$, then $(4U_5(105))^2 = 16U_5(105)=U_5(105)$. This shows that the order of $4U_5(105)$ is 2.

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 19

There are 10 elements in $\mathbb{Z}_{10}$ and 4 elements in $U(10)$. This means that their external direct product has 40 elements. We know 2 has order 5 in $\mathbb{Z}_{10}$, and we know $9^2=81$ which equals 1 mod 10, so this means that 9 has order 2 in $U(10)$. The order of $(2,9)$ in the direct product is the least common multiple of the orders of the individual elements, which means we have $$|(2,9)|=\text{lcm}(|2|,|9|) = \text{lcm}(5,2)=10.$$ We are creating a quotient group of a group of order 40, using a subgroup of order 10. This means the factor group has order $40/10 = 4$.

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 218-220 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 $2^3\cdot 7^2\cdot 3$.

In all the work below, we'll be working with external and/or internal direct products. Recall that we say $G$ is the internal direct product of two subgroups $H$ and $K$ if
  1. we know that $G=HK$ as a set product,
  2. we know that $H\cap K = \{e\}$, and
  3. we know that both $H$ and $K$ are normal.

Because we are working with Abelian groups, the third fact is immediate anytime we have subgroups. Anytime we know that $G$ is an internal direct product of $H$ and $K$, then we know that $$G=HK\approx H\times K = H\oplus K.$$ Some authors will just write $G=H\times K$ instead of $G\approx H\times K$, because the external direct product is equal to the internal direct product $HK$. I'll be sticking with the use of $\approx$, intead of just saying that $G$ equals its external direct product.

The next theorem shows that we can take any Abelian group and separate the group appart as $G=H_{p_1}\times H_{p_2}\times\cdots H_{p_k}$ where $p_1, p_2, \ldots, p_k$ are distinct primes, and every element in $H_{p_i}$ has a prime-powered order.

Problem.Splitting an Abelian Group into $H\times K$ where all elements of prime $p$ powered order are in $H$

Suppose that $G$ is an Abelian group with order $p^km$ where $p$ does not divide $m$. Let $H=\{x\in G\mid x^{p^k}=e\}$. Let $K=\{x\in G\mid x^m=e\}$.

  1. Prove that both $H$ and $K$ are normal subgroups of $G$, that $H\cap K=\{e\}$, and that $G=HK$. In other words, prove that $G$ is the internal direct product of $H$ and $K$, which means that $G\approx H\times K$.
  2. Further more, show that $|H|=p^k$ and $|K|=m$.

Now that we've split up the group $G$ into a direct sum of prime powered subgroups, we need to show that each prime powered subgroup can be written as a direct product of cyclic groups. To do this, we'll show that we can write $G$ in the form $G\approx \left<a\right>\times$ where $a$ has maximal order in $G$ and $K$ is a subgroup of $G$. Once we have shown that we can strip off one cyclic group from $G$, we just repeat this process until we obtain $$G\approx \left<a_1\right>\times\left<a_2\right>\times \left<a_3\right>\times\cdots\times \left<a_k\right> \times \left<e\right>$$ where the orders satisfy $|a_1|\geq |a_2|\geq |a_3|\geq \cdots \geq |a_k|>1$.

Problem.Obtaining two cyclic subgroups with trivial intersection in a prime-powered Abelian group, one of which has maximal order

Suppose that $G$ has order $p^n$ for some prime $p$ and positive integer $n$, and suppose that $G$ is not cyclic. Let $a$ be an element of $G$ with maximal order. Let $b\in G$ be an element of smallest order such that $b\notin \left<a\right>.$ Prove that $\left<a\right>\cap \left<b\right> = \{e\}$.

Problem.The Greedy Algorithm for a prime-powered Abelian group

Suppose that $G$ has order $p^n$ for some prime $p$ and positive integer $n$. Let $a$ be an element of $G$ with maximal order. Prove that $G\approx \left<a\right>\times K$ for some subgroup $K$ of $G$.

You'll need to use induction on the order of a group to complete this problem. The key is to use the subgroup $\left<b\right>$ from the previous problem and then examine the factor group $G/\left<b\right>$ which has a smaller order than $G$. You'll need to prove that the order of $a$ in $G$ is the same as the order of $\left<b\right>a$ in $G/ \left<b\right>$.

You'll also need to use the fact that subgroups of $G/ \left<b\right>$ correspond to subgroups of $G$ in a very nice way (something that we showed in the last two weeks).

Problem.A finite Abelian Group of prime power order is a direct product of cyclic groups

Suppose that $G$ has order $p^n$ for a prime $p$ and positive integer $n$. Prove that $G$ can be written as a direct product of cyclic groups.

Problem.The Uniqueness of a Direct Product Decomposition for prime-powered groups

Suppose that $G$ is a group of prime power order, and that $G$ has been written in two ways as the product of cyclic groups, namely that $G\approx H_1\times H_2\times \cdots \times H_n$ and $G\approx K_1\times K_2\times \cdots \times K_m$, where the subgroups are written so that their orders are nonincreasing, and none of the factors are trivial. Prove that $n=m$ and that $H_i\approx K_i$ for each $i$ from 1 to $n$.