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Thread: Finite abelian group.

  1. #1
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    Finite abelian group.

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

    Suppose $\displaystyle G_1 \cong G_2$. We have $\displaystyle G_1 \cong G_1(p_1) \times ...\times G_1(p_n)$ and $\displaystyle G_2 \cong G_2(p_1) \times ...\times G_1(p_n)$.
    But we know each of $\displaystyle G_1(p_i)$ and $\displaystyle G_2(p_i)$ can be decomposed into a product of cyclic groups. So, $\displaystyle G_1(p_i) \cong Z_{{p_i}^{l_1}} \times Z_{{p_i}^{l_2}} \times...\times Z_{{p_i}^{l_s}}$ where $\displaystyle Z_{x}$ is a cyclic group of order x.
    Also, $\displaystyle G_2(p_i) \cong Z_{{p_i}^{j_1}} \times Z_{{p_i}^{j_2}} \times...\times Z_{{p_i}^{j_s}}$
    So, I try to show that $\displaystyle Z_{{p_i}^{l_1}} \times Z_{{p_i}^{l_2}} \times...\times Z_{{p_i}^{l_s}} \cong Z_{{p_i}^{j_1}} \times C_{{p_i}^{j_2}} \times...\times Z_{{p_i}^{j_s}}$
    But I don't really know how to apply the hypothesis to prove this here.
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  2. #2
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    Quote Originally Posted by jackie View Post
    Can someone give me a hand on this problem? I want to show that $\displaystyle G_1 \cong G_2 \Leftrightarrow G_1(p_i) \cong G_2(P_i)$ for all $\displaystyle i \in \{1,2,...,n\}$. Given that $\displaystyle G_1, G_2$ are abelian groups such that $\displaystyle ord(G_1)=ord(G_2)={p_1}^{k_1}{p_2}^{k_2}...{p_n}^{ k_n}$

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

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

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

    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
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    Quote Originally Posted by jackie View Post
    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 $\displaystyle p_1$...
    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
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  5. #5
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    Quote Originally Posted by tonio View Post
    Supose first that [tex]G_1\,,\;G_2[tex] are cyclic, so that there's one single prime $\displaystyle p_1$...
    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 $\displaystyle G_1, G_2$ being cyclic. They might not be cyclic as well right because we only know that they are abelian.
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  6. #6
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    Here is my attempt to show one direction. Can someone help me if I'm on the right track.
    Instead of decomposing each $\displaystyle G_1(p_i)$ and $\displaystyle G_2(p_i)$ into direct products of cyclic subgroups, I tried to do induction on $\displaystyle i$
    I have $\displaystyle G_1 \cong G_1(p_1) \times ...\times G_1(p_n)$ and $\displaystyle G_2 \cong G_2(p_1) \times ...\times G_2(p_n)$
    For $\displaystyle i=1$, then $\displaystyle G_1 \cong G_(p_1)$ and $\displaystyle G_2 \cong G_2(p_1)$.
    By hypothesis, $\displaystyle G_1 \cong G_2$, so we can use transitivity of group isomorphism to get $\displaystyle G_1(p_1) \cong G_2(p_1)$
    Assume this is true for $\displaystyle i=n-1$, i.e $\displaystyle G_1(p_i) \cong G_2(p_i)$ for $\displaystyle i \in \{1,...,n-1\}$.
    I want to show this is true for $\displaystyle i=n$
    I have $\displaystyle G_1 \cong G_1(p_1) \times ...\times G_1(p_{n-1}) \times G_1(p_n) \cong G_2(p_1) \times ...\times G_2(p_{n-1}) \times G_2(p_n)$

    Here can I use the argument that if $\displaystyle G \times K \cong H \times K$ where $\displaystyle G, H, K$ are finite abelian, then $\displaystyle G \cong H$ to say that $\displaystyle G_1(p_n) \cong G_2(p_n)$? I appreciate any help.
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  7. #7
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    It seems that my attempt to use induction is completely false because I can't assume $\displaystyle G_1 \cong G_1(p_1)$ and $\displaystyle G_2 \cong G_2(p_1)$ My TA suggests that I should try to look for isomorphism from $\displaystyle G_1(p_i) \rightarrow G_2(p_i)$ Anyone can help me out here please.
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