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Problem: If $G$ has exactly one subgroup $H$ of order $k$, then $H$ is normal in $G$.

I can see the obvious low level solution, that $gHg^{-1}$ is also a subgroup of order $k$ and therefore $H$ is normal. What confuses me is that this problem was in the chapter titled "Homomorphisms". The author introduces the idea of a homomorphism, proves the isomorphism theorems, and then hands us this problem.

I was thinking to try to use the first isomorphism in an obvious way such that $H$ is the kernel of some homomorphism but I couldn't find it.

Can someone help me find the likely "intended" solution? Since the purpose of exercises should be to help me understand homomorphisms and ismorphisms better, my solution can't really be considered satisfactory.

Matt Samuel
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Nitin
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  • This question, when we read the title, is a duplicate as noted. However when we read the body of the question there are other things. It asks why this question is called an exercise in homomorphism. I'd not call the a duplicate after reading the question in entirety. – P Vanchinathan Oct 27 '16 at 04:06

2 Answers2

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Inner automorphisms are homomorphisms. Image of a subgroup under any homomorphism is again a subgroup. You are using this result about homomorphisms in your proof. So if it is an automorphism order is also preserved. For this reason that problem is listed under homomorphisms, Justifiably so.

P Vanchinathan
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P Vanchinathan's answer is correct. For some additional advice, don't read too much into the phrasing or placement of an exercise. It can give you perhaps a hint as to where to look in your mind for a proof, but no more than that. An intended answer is no more or less correct than an unintended correct answer, provided your proof uses only the material introduced up to that point. If you're taking an exam and there's a question where the intended answer uses material you didn't study, you will get full credit for finding an alternative proof using only the material you did study.

Matt Samuel
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