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Math Help - group element order

  1. #1
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    group element order

    Let G be an abelian group and let T={a in G|a^m=e for some m>1}. Prove that T is a subgroup of G and that G/T has no element - other than its identity element - of finite order.

    Please show details of this proof. It is another practice problem for my final exam.

    Thank you so much!
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  2. #2
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    Quote Originally Posted by mpryal View Post
    Let G be an abelian group and let T={a in G|a^m=e for some m>1}. Prove that T is a subgroup of G and that G/T has no element - other than its identity element - of finite order.

    Please show details of this proof. It is another practice problem for my final exam.

    Thank you so much!
    The first thing we should do is try to prove closure. Choose a,b\in T. Then there exists positive integers m,n with a^m=e and b^n=e, with m,n>1. Since a,b\in G, then ab\in G. But is it also in T? Well, observe that (ab)^{mn}\in G. Also, (ab)^{mn}=a^{mn}b^{mn}=(a^m)^n(b^n)^m=e^ne^m=e. So then the conditions are satisfied such that ab\in T, and therefore T is closed.

    It is pretty obvious that e\in T.

    It's almost just as obvious that for any a\in T then a^{-1}\in T. But let's prove it anyway. There exists integer m>1 with a^m=e. Observe also that since T is abelian, we can do the following: e=a^{-1}a=(a^{-1}a)^m=(a^{-1})^ma^m=(a^{-1})^m. So for any a\in T, then a^{-1}\in T.

    That should do it.
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  3. #3
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by mpryal View Post
    Let G be an abelian group and let T={a in G|a^m=e for some m>1}. Prove that T is a subgroup of G and that G/T has no element - other than its identity element - of finite order.

    Please show details of this proof. It is another practice problem for my final exam.

    Thank you so much!
    Hi mpryal.

    Let gT\in G/T and suppose (gT)^k=e_{G/T}=T for some k\in\mathbb Z^+. Then g^kT=T and so g^k\in T and so (g^k)^m=g^{km}=e for some k\in\mathbb Z^+. Hence g is of finite order in G, i.e. g\in T so that gT=T. This shows that T is the only element of G/T of finite order.
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