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Thread: Group Homomorphism?

  1. #1
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    Group Homomorphism?

    Let G,G' be groups and φ:G-->G' be a homomorphism. Then prove:
    a)if o(g)=n ,then o(φ(g))|n
    b)if o(g)=n and φ is injective, then o(φ(g))=n

    Firstly I don't actually understand the question. What does the o mean?Is it a new mapping which hasn't been defined or does it represent the composition of functions and if so then how is it computed?Secondly does the |n in (a) represent "divides n" and if so how is that proved, if not then what does it represent? Neither of these appears in my abstract algebra text. In (b) I know that an injective function is one to one but again I don't really understand the question.

    Any help would be greatly appreciated. Thanks in advance.
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  2. #2
    Senior Member Tinyboss's Avatar
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    Re: Group Homomorphism?

    o(g) refers to the order of g, which is the smallest positive integer n such that g^n is the identity, or infinity if no such n exists.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Re: Group Homomorphism?

    Quote Originally Posted by chocaholic View Post
    Let G,G' be groups and φ:G-->G' be a homomorphism. Then prove:
    a)if o(g)=n ,then o(φ(g))|n
    b)if o(g)=n and φ is injective, then o(φ(g))=n

    Firstly I don't actually understand the question. What does the o mean?Is it a new mapping which hasn't been defined or does it represent the composition of functions and if so then how is it computed?Secondly does the |n in (a) represent "divides n" and if so how is that proved, if not then what does it represent? Neither of these appears in my abstract algebra text. In (b) I know that an injective function is one to one but again I don't really understand the question.

    Any help would be greatly appreciated. Thanks in advance.
    Yes, $\displaystyle \mid$ means divide. I'll give you the one bit of information that is needed for this problem, namely if $\displaystyle g^k=e$ then $\displaystyle |g|\mid k$ (where I use $\displaystyle |\cdot|$ instead of $\displaystyle o(\cdot)$). From this I think a) should be pretty immediate. For the second, try doing the opposite of the technique you do for the first by proving that $\displaystyle n\mid |\varphi(g)|$.
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