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Let H be a subgroup of a group G. Let $k,g \in G$ such that $gH = Hk$. Suppose further that $[G:H]$ is a prime integer, and that $g \notin H$. Prove that H is normal in G.

I have totally no idea at all how to do this question. Can someone tell me how to start?

macho
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    Your question is not clear. What is the connection of the $k$ and $g$ with your question? – Babai Nov 25 '14 at 15:08
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    I guess you want to say for every $g\notin H$ there exists $k\in G$ s.t $gH=Hk$ – Babai Nov 25 '14 at 15:09
  • @Susobhan I rather think the assumption is that there exist $g,k\in G$ with $g\notin H$, such that $gH = Hk$, for these two elements. It would be trivial, and the assumption that the index is prime irrelevant if it were $\forall g \exists k$. Can you confirm that, macho? – Daniel Fischer Nov 25 '14 at 15:40
  • It seems to me that the question is: if every left coset of $H$ is a right coset and if the index is prime, then $H$ is normal. – lhf Nov 25 '14 at 15:53
  • sorry i typed wrongly.Edited – macho Nov 25 '14 at 16:03
  • Whatever you have edited macho , still doesnt make much sense. it looks the same. – Babai Nov 25 '14 at 16:17

1 Answers1

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If there are $g,k\in G$ with $g\notin H$ and $gH = Hk$, then for this particular $g$ we have $gH = Hg$, or equivalently $gHg^{-1} = H$. This follows since under these hypotheses $g \in Hg \cap Hk = Hg \cap gH$, and two right cosets are either disjoint or equal.

Thus for the normalizer $$N_G(H) = \{x\in G : xHx^{-1} = H\} \tag{$\ast$}$$ we know $g \in N_G(H)$, and consequently $H \subsetneqq N_G(H)$. Since $$[G:H] = [G : N_G(H)] \cdot [N_G(H) : H]$$ is prime and $[N_G(H) : H] > 1$ by $(\ast)$ it follows that $[G : N_G(H)] = 1$, i.e. $N_G(H) = G$, in other words $H$ is normal.

Daniel Fischer
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  • what does "Let G act by conjugation on the set of conjugates of H." mean? – macho Nov 25 '14 at 16:13
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    Consider the set $A = { xHx^{-1} : x \in G}$. That is the set of conjugates of $H$. On that set, let $G$ act by the mapping $\kappa\colon (y, xHx^{-1}) \mapsto y(xHx^{-1})y^{-1}$. Since the latter is $(yx)H(yx)^{-1}$, for every $y\in G$ the map $\kappa_y \colon M \mapsto yMy^{-1}$ is a permutation of $A$. – Daniel Fischer Nov 25 '14 at 16:18
  • is this group action? Is it possible to solve without using group action because I have not learnt that – macho Nov 25 '14 at 16:47
  • Yes, we can do it without group actions. I've edited the answer accordingly. – Daniel Fischer Nov 25 '14 at 16:55
  • To continue the proof, can we say that $N_G(H)$ is a subgroup of $G$ containing $H$. So $|H|$ divides $|N_G(H)|$ divides $|G|=p|H|$. So $N_G(H)$ is either $H$ or $G$. Since $g \notin H$, the former is not possible. Hence the latter is true? – eatfood Nov 15 '18 at 13:20
  • For anyone else who got stuck (like me) at showing $gH = Hg$: Note that $g \in gH = Hk$, hence there exists some $h \in H$ such that $g = hk$. Then $Hg = H(hk) = Hk = gH$ as required... – eatfood Sep 27 '20 at 14:58
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    @eatfood Or we can use $g \in Hg \cap Hk$, whence the two right cosets are the same. And regarding your previous comment (which came while I was absent from the site, so I didn't see it until now): yes, that's the argument. Since $g \in N_G(H)$ we have $H \subsetneqq N_G(H)$, hence $[N_G(H) :H] > 1$, and since $[G:H]$ is prime it follows that $N_G(H) = G$. Thanks for commenting here, now I can un-hint the answer. – Daniel Fischer Sep 27 '20 at 15:08