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Math Help - 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 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.
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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 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
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  3. #3
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    Quote Originally Posted by tonio View Post
    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
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  4. #4
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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 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 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.
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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 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.
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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 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.
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