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Math Help - Isomorphic groups

  1. #1
    Super Member Deadstar's Avatar
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    Isomorphic groups

    Trying to brush up on my group theory before next year...

    One thing I've always had trouble with is proving two groups are (or in this case aren't) isomorphic.

    So here's my Q.

    Show that C_8 and C_4 \times C_2 are not isomorphic.

    My thoughts... I'm thinking something along the lines of, an element of order 6 in the group C_4 \times C_2 is just the identity element since if you have g \in C_4 \times C_2 then g^6 = g^{4+2} = g^4 \cdot g^2 = e \cdot e = e? Whereas an element of order 6 in the group C_8 is not.

    Think this is wrong so any help appreciated.
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Deadstar View Post
    Trying to brush up on my group theory before next year...

    One thing I've always had trouble with is proving two groups are (or in this case aren't) isomorphic.

    So here's my Q.

    Show that C_8 and C_4 \times C_2 are not isomorphic.

    My thoughts... I'm thinking something along the lines of, an element of order 6 in the group C_4 \times C_2 is just the identity element since if you have g \in C_4 \times C_2 then g^6 = g^{4+2} = g^4 \cdot g^2 = e \cdot e = e? Whereas an element of order 6 in the group C_8 is not.

    Think this is wrong so any help appreciated.

    There are no elements of order 6 in either group - let G be an arbitrary group of order n, and let g \in G be any element in G. Then o(g) | |G| - that is, the order of any element of a finite group divides the order of the group.

    Hint: prove that for (g,h) \in C_n \times C_m, then o((g,h)) = max(o(g), o(h)).
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  3. #3
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by Swlabr View Post
    Hint: prove that for (g,h) \in C_n \times C_m, then o((g,h)) = max(o(g), o(h)).
    That is incorrect. You want the lcm, not the max.

    @Deadstar: C_8 has an element of order 8. What is the maximum order of any element in C_4\times C_2\,? Use Swlabrís hint (corrected as above).
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by TheAbstractionist View Post
    That is incorrect. You want the lcm, not the max.
    Thanks - I should be less hasty with my responses.
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