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A problem from Artin's Algebra (not the problem I am asking):

Let $G$ be an abelian group of odd order. Prove that the map $\varphi : G\rightarrow G$ defined by $\varphi (x)=x^2$ is an automorphism.

I have solved this problem, also we can see here.

Then comes Generalization of the problem:

Let $G$ be an abelian group of finite order. Prove that the map $\varphi : G\rightarrow G$ defined by $\varphi (x)=x^k$ is an automorphism, where $k$ does not have any prime factor same of order of $G$.

Is it ok?

user1729
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MAN-MADE
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  • Doesn't the same argument go through with very few changes? – lulu Jul 13 '17 at 13:44
  • @lulu I just started group theory, so can you please point out the changes. – MAN-MADE Jul 13 '17 at 13:46
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    Hints: as with the first case, we need to show the kernel is trivial. So suppose $g\neq e$ is in the kernel. Let $o$ be the order of $g$. Then $o$ must divide $|G|$ and it must divide $k$. – lulu Jul 13 '17 at 13:47

1 Answers1

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Here is a roadmap:

  • Prove that $\varphi$ is a homomorphism. Use that $G$ is abelian.

  • Prove that $\varphi$ is injective. Use that the order of $G$ is odd.

  • Prove that $\varphi$ is surjective. Use that $G$ is finite.

For the generalization, use that $\gcd(k,n)=1$ implies $ku+nv=1$ for $u,v \in \mathbb Z$ in the second step.

lhf
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  • $n=|G|$. Thanks. I can do it now. – MAN-MADE Jul 13 '17 at 14:18
  • @MANMAID, for extra credit, prove that $\varphi$ is still a bijection, even when $G$ is not abelian. Hint: use the abelian case. – lhf Jul 13 '17 at 14:20
  • Sir, I could not do that you mentioned(non-abelian part). Surjective part is same. In injective part I proceeded in this way: Since $|G|$ is odd, we must get a subgroup of abelian group in $G$. Let it be $A$. Let $x,y\in A$, $x^2=y^2\Rightarrow x=y$. $(ax)^2=(by)^2\Rightarrow (ab^{-1})^{-1}x(ab^{-1})=x$. What will I do next? – MAN-MADE Jul 13 '17 at 16:09
  • I don't think this will help! – MAN-MADE Jul 13 '17 at 16:17
  • @MANMAID, you're right. In the non-abelian case you prove the surjective part first, by considering cyclic subgroups. See https://math.stackexchange.com/a/828662/589. – lhf Jul 13 '17 at 16:58
  • Sir, from your reputation and score in abstract algebra, I think you are very good at abstract algebra. I am asking: I started abstract algebra from Artin's book, is it ok or there is much better book for abstract algebra which covers conceptual part more rigorously. (I don't want "easy to understand books" or very "elementary books", since I have completed masters in Statistics and thinking of study mathematics rigorously, for my keen interest in maths :) ) [sorry to bother you, I won't do that again!] – MAN-MADE Jul 19 '17 at 14:36
  • @MANMAID, I suggest you ask a separate question on books, after reviewing the question on books already posted here. If you decide to post a question, be sure to include enough detail about your expectations of a book and why the ones you have read do not fulfill them. – lhf Jul 19 '17 at 14:50
  • @MAN-MADE How is your algebra reading going on? – Babai Jun 28 '18 at 19:10