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Math Help - 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, \mid means divide. I'll give you the one bit of information that is needed for this problem, namely if g^k=e then |g|\mid k (where I use |\cdot| instead of 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 n\mid |\varphi(g)|.
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