1. ## Finite abelian group.

Can someone give me a hand on this problem? I want to show that $G_1 \cong G_2 \Leftrightarrow G_1(p_i) \cong G_2(P_i)$ for all $i \in \{1,2,...,n\}$. Given that $G_1, G_2$ are abelian groups such that $ord(G_1)=ord(G_2)={p_1}^{k_1}{p_2}^{k_2}...{p_n}^{ k_n}$

Suppose $G_1 \cong G_2$. We have $G_1 \cong G_1(p_1) \times ...\times G_1(p_n)$ and $G_2 \cong G_2(p_1) \times ...\times G_1(p_n)$.
But we know each of $G_1(p_i)$ and $G_2(p_i)$ can be decomposed into a product of cyclic groups. So, $G_1(p_i) \cong Z_{{p_i}^{l_1}} \times Z_{{p_i}^{l_2}} \times...\times Z_{{p_i}^{l_s}}$ where $Z_{x}$ is a cyclic group of order x.
Also, $G_2(p_i) \cong Z_{{p_i}^{j_1}} \times Z_{{p_i}^{j_2}} \times...\times Z_{{p_i}^{j_s}}$
So, I try to show that $Z_{{p_i}^{l_1}} \times Z_{{p_i}^{l_2}} \times...\times Z_{{p_i}^{l_s}} \cong Z_{{p_i}^{j_1}} \times C_{{p_i}^{j_2}} \times...\times Z_{{p_i}^{j_s}}$
But I don't really know how to apply the hypothesis to prove this here.

2. Originally Posted by jackie
Can someone give me a hand on this problem? I want to show that $G_1 \cong G_2 \Leftrightarrow G_1(p_i) \cong G_2(P_i)$ for all $i \in \{1,2,...,n\}$. Given that $G_1, G_2$ are abelian groups such that $ord(G_1)=ord(G_2)={p_1}^{k_1}{p_2}^{k_2}...{p_n}^{ k_n}$

Suppose $G_1 \cong G_2$. We have $G_1 \cong G_1(p_1) \times ...\times G_1(p_n)$ and $G_2 \cong G_2(p_1) \times ...\times G_1(p_n)$.
But we know each of $G_1(p_i)$ and $G_2(p_i)$ can be decomposed into a product of cyclic groups. So, $G_1(p_i) \cong Z_{{p_i}^{l_1}} \times Z_{{p_i}^{l_2}} \times...\times Z_{{p_i}^{l_s}}$ where $Z_{x}$ is a cyclic group of order x.
Also, $G_2(p_i) \cong Z_{{p_i}^{j_1}} \times Z_{{p_i}^{j_2}} \times...\times Z_{{p_i}^{j_s}}$
So, I try to show that $Z_{{p_i}^{l_1}} \times Z_{{p_i}^{l_2}} \times...\times Z_{{p_i}^{l_s}} \cong Z_{{p_i}^{j_1}} \times C_{{p_i}^{j_2}} \times...\times Z_{{p_i}^{j_s}}$
But I don't really know how to apply the hypothesis to prove this here.

As $G_1\cong G_2$ they both have exactly the same elements of every single order dividing $|G_1|=|G_2| \Longrightarrow$ apply induction to show $C_{{p_i}^{l_1}} \times C_{{p_i}^{l_2}} \times...\times C_{{p_i}^{l_s}} \cong C_{{p_i}^{j_1}} \times C_{{p_i}^{j_2}} \times...\times C_{{p_i}^{j_s}}$

Tonio

3. Originally Posted by tonio
As $G_1\cong G_2$ they both have exactly the same elements of every single order dividing $|G_1|=|G_2| \Longrightarrow$ apply induction to show $C_{{p_i}^{l_1}} \times C_{{p_i}^{l_2}} \times...\times C_{{p_i}^{l_s}} \cong C_{{p_i}^{j_1}} \times C_{{p_i}^{j_2}} \times...\times C_{{p_i}^{j_s}}$

Tonio
Thanks a lot for your help Tonio, but can you elaborate a bit on the induction here because I don't see what I need to induct on. So I have no idea what the base case is to proceed. I'm slow on this

4. Originally Posted by jackie
Thanks a lot for your help Tonio, but can you elaborate a bit on the induction here because I don't see what I need to induct on. So I have no idea what the base case is to proceed. I'm slow on this
Supose first that [tex]G_1\,,\;G_2[tex] are cyclic, so that there's one single prime $p_1$...
After that, assume for n-1 primes and prove for n primes, and what you'll do is take quotient rings to reduce the number of primes...

Tonio

5. Originally Posted by tonio
Supose first that [tex]G_1\,,\;G_2[tex] are cyclic, so that there's one single prime $p_1$...
After that, assume for n-1 primes and prove for n primes, and what you'll do is take quotient rings to reduce the number of primes...

Tonio
So, do I have to do two cases here because you said to consider $G_1, G_2$ being cyclic. They might not be cyclic as well right because we only know that they are abelian.

6. Here is my attempt to show one direction. Can someone help me if I'm on the right track.
Instead of decomposing each $G_1(p_i)$ and $G_2(p_i)$ into direct products of cyclic subgroups, I tried to do induction on $i$
I have $G_1 \cong G_1(p_1) \times ...\times G_1(p_n)$ and $G_2 \cong G_2(p_1) \times ...\times G_2(p_n)$
For $i=1$, then $G_1 \cong G_(p_1)$ and $G_2 \cong G_2(p_1)$.
By hypothesis, $G_1 \cong G_2$, so we can use transitivity of group isomorphism to get $G_1(p_1) \cong G_2(p_1)$
Assume this is true for $i=n-1$, i.e $G_1(p_i) \cong G_2(p_i)$ for $i \in \{1,...,n-1\}$.
I want to show this is true for $i=n$
I have $G_1 \cong G_1(p_1) \times ...\times G_1(p_{n-1}) \times G_1(p_n) \cong G_2(p_1) \times ...\times G_2(p_{n-1}) \times G_2(p_n)$

Here can I use the argument that if $G \times K \cong H \times K$ where $G, H, K$ are finite abelian, then $G \cong H$ to say that $G_1(p_n) \cong G_2(p_n)$? I appreciate any help.

7. It seems that my attempt to use induction is completely false because I can't assume $G_1 \cong G_1(p_1)$ and $G_2 \cong G_2(p_1)$ My TA suggests that I should try to look for isomorphism from $G_1(p_i) \rightarrow G_2(p_i)$ Anyone can help me out here please.