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I'm trying to show that any subgroup $H$ of a group $G$ having finite index must contain a normal subgroup of $G$ of finite index.

I tried to define a homomorphism $\psi:G/H \to G/H$ given by $\psi(xH) =gxH$ and prove that $\ker(\psi)\subset H$. Is there another way to solve this problem?

carmichael561
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clm
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    That strategy sounds good. Where did you get stuck with it? – stewbasic May 25 '17 at 23:23
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    The concept of "a homomorphism $\psi : G/H \to G/H$" is nonsense unless you already know that $H$ is normal, because if $H$ is not normal then $G/H$ is not a group. To be more precise, $G$ does not induce a well-defined group structure on the set of right cosets $G/H$ unless $H$ is normal. – Lee Mosher May 26 '17 at 00:13
  • It's not "quite" the same question, as there is no stipulation that the normal subgroup we seek must lie within $H$. I'm not opposed to the close vote, however. – David Wheeler May 26 '17 at 01:22
  • @DavidWheeler: Then it is a duplicate of https://math.stackexchange.com/questions/494207/if-h-leq-g-is-of-finite-index-then-we-have-a-normal-subgroup-n-leq-g-of-fi?rq=1 or https://math.stackexchange.com/questions/2085531/if-h-leq-g-has-finite-index-then-g-has-a-normal-subgroup-of-finite-index?rq=1 – Moishe Kohan May 26 '17 at 01:36
  • It happens. I would agree it is an exact duplicate of the first, and the second link you provide duplicates the first possible duplicate. You might successfully get it closed. I am ambivalent. – David Wheeler May 26 '17 at 03:50

3 Answers3

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Given a subgroup $H\leq G$ of finite index, $G$ acts transitively on the set of left cosets $G/H$ via $g\cdot(g'H) := (gg')H$, for $g \in G$ and $g'H \in G/H$; the kernel $K$ of the corresponding group homomorphism $G \to \text{Sym}(G/H)$ is a normal subgroup of $G$ contained in $H$, and hence has finite index.

Comment: One can show that $K = \bigcap_{g \in G} gHg^{-1}$. This is core of $H$ in $G$; it is the largest normal subgroup of $G$ contained in $H$.

cat
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Hint: Consider the induced homomorphism from $G$ to the group of permutations of $G/H$.

tomasz
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    shouldn't short hints like this be in the comments and not answers – JJR May 26 '17 at 00:09
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    @JJR: Why? Anything longer than that will spoil the answer, in my opinion. Since this looks very much like a homework problem or a book exercise, that would be beside the point. – tomasz May 26 '17 at 00:11
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    This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review – The Count May 26 '17 at 00:17
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    @TheCount I agree with tomasz, any more would spoil the question. The hint is perfect. (Although one could mention along with the others that OP's idea of a homomorphism $G/H\to G/H$ won't work.) – anon May 26 '17 at 01:43
  • @arctictern As I always say when someone responds to a review comment: the item came up in my queue and I reviewed it as I saw fit. You are free to disagree. – The Count May 26 '17 at 01:48
  • Duh. ${}{}{}{}$ – anon May 26 '17 at 01:52
  • No need to be rude, @arctictern. What sort of response did you expect? And ping me if you have something to say! – The Count May 26 '17 at 14:46
  • @TheCount In fact, I thought your response was amusing: what I expected was either an explanation or no response, but you gave a response with no explanation! Doesn't matter; people don't need reasons, and don't need to explain themselves to others when they do. As for rude, consider what it looks informing a $5$-year member that MSE has a comment system and insinuating they need clarification from OP... Heh. Although, true, "duh" was terse and overly pessimistic about the prospect of engaging you, sorry. – anon May 26 '17 at 16:12
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    At any rate, providing hints to questions has long been an accepted, commonplace tradition on MSE. On the one hand, it has pedagogical value, and on the other hand it is a compromise between users' desire to help sincere askers and the community's attitude against the "do my homework for me" tide. If you're interested you can create (another) meta thread about hints, advocating (say) a more legalistic view of not submitting hints in the answer box because the word "answer" is in the phrase "Your Answer." – anon May 26 '17 at 16:12
  • @arctictern How long you've been here is irrelevant. And I have no problem with you, you just helped me with a problem the other day. But as there is no universally agreed upon or enforced rule for such small hints, opinion reigns. I'm not going to change that. – The Count May 26 '17 at 16:18
  • @TheCount Whether or not your original comment to tomasz was rude does depend on how long tomasz has been here. You (or really, the template you probably clicked instead of writing a response yourself) explains MSE affairs to him like he's new and ignorant, which is very awkward to do to someone who is perfectly knowledgeable about how MSE works. tomasz is clearly not trying to critique or request clarification. Indeed, if you'd written the comment yourself as-is instead of selected it, I'd find it almost passive aggressive. – anon May 26 '17 at 16:34
  • I think I've bothered tomasz enough, so I'll leave it at this: if you don't want to engage me or the community about why you think hints are bad for the site in every and all cases, then you don't have to, but please don't ask us to engage you if you're not going to engage us. – anon May 26 '17 at 16:34
  • @arctictern I'm sure you do the review queues and know how they work since you've been here so long. I clicked "does not answer the question" and the comment is auto-generated. Your problem is with the tone of an automatic message I have no control over. And I have engaged just as often as you have, while quibbling over what I think is a tiny little insignificant point about a comment that didn't even have anything to do with you. It is just silly to respond to a review queue comment. – The Count May 26 '17 at 16:41
  • @arctictern: Thanks for defending me so vigorously. I think your comment: "At any rate, providing hints to questions has long been an accepted, commonplace tradition on MSE. On the one hand, it has pedagogical value, and on the other hand it is a compromise between users' desire to help sincere askers and the community's attitude against the "do my homework for me" tide." leaves nothing for me to add. As for The Count's comment, I would maybe find it rude, but as he said, this is an automatically generated comment, so if anything, I find it kind of funny. – tomasz May 27 '17 at 19:07
  • @TheCount: while I agree that arctictern is overreacting on the rudeness front (I don't feel offended, even though I strongly disagree with your assesment for reasons explained by arctictern, as well as my previous comments), I don't think it's silly to reply to comments from the review queue. If the comments were not meant to be replied to, they would not be comments at all. – tomasz May 27 '17 at 19:10
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As another answer indicates, you're "close".

The problem is, the map $\psi_g: G/H \to G/H$ given by $\psi_g(xH) = (gx)H$ might not be a homomorphism, because $G/H$ might not be a group.

So is there a way to fix this?

It turns out there is, and it's common to different branches of algebra.

The map $\psi_g$ isn't necessarily a homomorphism, but it is bijective on $G/H$, for if:

$\psi_g(xH) = \psi_g(yH)$, then $(gx)H = (gy)H$, that is: $(gy)^{-1}(gx) \in H$.

Expanding this, we have: $y^{-1}g^{-1}gx = y^{-1}x \in H$, so that $xH = yH$.

This shows that $\psi_g$ is injective, and since $G/H$ is finite, it is thus also bijective. And we know the bijections on any set (even $G/H$) form a group.

So now we have two groups, $G$ and $\text{Sym}(G/H)$. So we can ask: can we find a homomorphism now?

Your task, now, is to convince yourself (this means prove) that $g \mapsto \psi_g$ is the homomorphism we're after. Let's call this homomorphism $\phi$ (so that $\phi(g) = \psi_g$).

Now, to finish, consider the two main objects of interest with our homomorphism $\phi$-namely, its kernel and image. Fill in the blanks:

$\text{im }\phi$ is a subgroup of a ______ group, so it is _____ .

Therefore, the kernel has ______ ______ (via the Fundamental Isomorphism Theorem).

One last question for you: have we found a suitable normal subgroup?

David Wheeler
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