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Math Help - External Direct Product

  1. #1
    Senior Member slevvio's Avatar
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    External Direct Product

    Let C_n denote a cyclic group of order n. Show that if p is a prime number then every abelian group of order p^2 is isomorphic to one of the two groups  C_{p^2}, C_p \times C_p .

    OK!

    Let G be an abelian group, with  |G|=p^2. Suppose G is not isomorphic to a cyclic group of order  p^2 . So we need to show that  G \cong C_p \times C_p.

    Let  g \in G with  g \not= 1_G . So |g| = p. |g| cannot be  p^2 because then g would generate the whole group and so G would be equal and isomorphic to a cyclic subgroup of order  p^2 .

    Let  h \in G \setminus <g>. Then |h| = p.

    Define  \theta : <g>\times <h> \rightarrow G, given by  \theta (g^i,h^j) = g^i g^j.

    This is a homomorphism (need to use abelian-ness to prove this)

    But how do I show that this is surjective and injective? It says here that  |<g> \times <h> | = |G| = p^2 , so it is only required to prove injectivity or surjectivity, but why is  |<g> \times <h> | = |G| = p^2 true?

    In trying to work out the kernel it seems that it is only equal to {1} if  <g> \cap <h> = \{1\} but is this true? Any help would be appreciated here. Thanks very much
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  2. #2
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    Quote Originally Posted by slevvio View Post
    Let C_n denote a cyclic group of order n. Show that if p is a prime number then every abelian group of order p^2 is isomorphic to one of the two groups  C_{p^2}, C_p \times C_p .

    OK!

    Let G be an abelian group, with  |G|=p^2. Suppose G is not isomorphic to a cyclic group of order  p^2 . So we need to show that  G \cong C_p \times C_p.

    Let  g \in G with  g \not= 1_G . So |g| = p. |g| cannot be  p^2 because then g would generate the whole group and so G would be equal and isomorphic to a cyclic subgroup of order  p^2 .

    Let  h \in G \setminus <g>. Then |h| = p.

    Define  \theta : <g>\times <h> \rightarrow G, given by  \theta (g^i,h^j) = g^i g^j.

    This is a homomorphism (need to use abelian-ness to prove this)

    But how do I show that this is surjective and injective? It says here that  |<g> \times <h> | = |G| = p^2 , so it is only required to prove injectivity or surjectivity, but why is  |<g> \times <h> | = |G| = p^2 true?

    In trying to work out the kernel it seems that it is only equal to {1} if  <g> \cap <h> = \{1\} but is this true? Any help would be appreciated here. Thanks very much

    If h^a=g^b , for some 0\leq a,b< p , then as (a,p)=1 there exists k\in\mathbb{Z} s.t. ak=1\!\!\!\pmod p\Longrightarrow h=h^{ak}=g^{bk}\in <g> , contradicting your choice of h , and
    this simply means <g>\cap <h>=1 .

    Tonio
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by slevvio View Post

    But how do I show that this is surjective and injective? It says here that  |<g> \times <h> | = |G| = p^2 , so it is only required to prove injectivity or surjectivity, but why is  |<g> \times <h> | = |G| = p^2 true?
    Because \left|\left\langle g\right\rangle\right|=\left|\left\langle h\right\rangle\right|=p. So from basic set theory and combinatorics \text{card }M=m,\text{card }N=n\implies \text{card }M\times n=mn. Also, the reason you only have to prove either surjectivity or injectivity is that a surjective function from a set to an equipotent set is injective and vice versa if the sets are finite.
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