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Let $G$ be a finite group and $H$ a subgroup of $G$ of index $n$, i.e., $[G:H]=n$. Prove that $$\forall g \in G,\; g^{n!} \in H.$$

This is a question I've had in a past exam for Group Theory and I'm really struggling to come up with a solution. There were two hints given in the question, namely that $\# G = n \cdot \# H$ (which is just from Lagrange's Theorem), and we were told to recall that $\#S_{n} = n!$.

I've thought about using Cayley's theorem, that every group is isomorphic to a subgroup of $S_{n}$, but can't figure out how to get the desired result.

user26857
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Daniel Ward
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  • For what values of $m$ we have $g^mH=H$? –  Jul 19 '16 at 14:15
  • See also http://math.stackexchange.com/questions/573050/if-h-is-a-subgroup-of-g-of-finite-index-n-then-under-what-condition-gn. – lhf Jul 19 '16 at 15:50

3 Answers3

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Alternatively, one can argue by contradiction.

So suppose there is a $g\in G$ with the property $g^{n!}\notin H$.

Then neither of the elements $g,g^{2},...,g^{n}$ can belong to $H$. This implies that $$H, Hg, Hg^{2},...,Hg^{n}$$ are distinct Right-cosets. But this violates the assumption $[G:H]=n$.

TheOscillator
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    Brilliant, kudos. This also shows clearly that the results holds with $n!$ replaced by the lcm of $1,2,\dots,n$, which is indeed the exponent of $S_n$. – Andreas Caranti Jul 21 '16 at 16:55
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Do it in two steps.

First you prove that there is normal subgroup $K$ of $G$ that is contained in $H$, and has index a divisor of $n!$

For this, as in the previous answer, you let $G$ act on the left cosets of $H$ by multiplication, regard this action as a homomorphism $G \to S_{n}$, and consider the kernel $K$.

Then you consider the factor group $G/K$, and use the fact that if a group $L$ has order $m$, then $g^{m} = 1$ for each $g \in L$.

Andreas Caranti
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  • Sorry, I'm not too confident on the topic of group actions. I've been told they're homomorphisms that map from the group to the set of permutations of the set (that the group is acting on). Is this correct? If so, why do we form a group action from $G$ to $S_{n}$? – Daniel Ward Jul 19 '16 at 14:30
  • I've just seen that in the question we've defined $[G:H]=n$. – Daniel Ward Jul 19 '16 at 14:32
  • Yes, that's it. You have the set of $n$ right cosets, say, of $H$ in $G$, and an element $g$ acts as a permutation on this set by $Ha \mapsto H a g$. If you associate to $g \in G$ this permutations, you get a homomorphism $G \to S_{n}$. – Andreas Caranti Jul 19 '16 at 14:34
  • Okay, so the kernel of (what I'm calling) $\varphi$ is going to be the elements of $G$ that map to the identity permutation. So is this going to be $g \in G$ s.t. $g \in Ha$ for each right coset? Or should it be more general than that, and should it hold regardless of which coset $g$ is in? – Daniel Ward Jul 19 '16 at 14:41
  • The kernel is $K = { g \in G : H a g = H a\ \text{for all $a \in G$} } = { g \in G : H a g a^{-1} = H\ \text{for all $a \in G$} } = { g \in G : a g a^{-1} \in H a\ \text{for all $a \in G$} } = { g \in G : g \in a^{-1} H a\ \text{for all $a \in G$} } = \bigcap_{a \in G} a^{-1} H a$. But what's important is that $K$ is normal in $G$, and $G/K$ is isomorphic to a subgroup of $S_{n}$. – Andreas Caranti Jul 19 '16 at 14:52
  • I see. So then because $K$ is normal we say $G/K$ is isomorphic to a subgroup of $S_{n}$, so it has order dividing $n!$. Therefore $\forall a \in G, \ (Ka)^{n!} = K \implies \forall a \in G, \ K(a^{n!}) = K \implies \forall a \in G, \ a^{n!} \in K \subseteq H$. Is that correct? – Daniel Ward Jul 19 '16 at 15:04
  • That's it, exactly! – Andreas Caranti Jul 19 '16 at 15:57
  • Sorry can I ask one more thing? I've revisited this today but now can't prove to myself that $K \subseteq H$ holds. I've tried to let some $g \in K \implies g = aha^{-1}$ for some $a \in G$, but I don't see that this is always in $H$. Am I missing something? – Daniel Ward Jul 20 '16 at 12:14
  • Look at the formula I wrote above. $K$ is the intersection of all conjugates of $H$, and $H$ is one of them. – Andreas Caranti Jul 21 '16 at 19:13
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Hint Consider the left cosets $X=G/H$ and the group action of $G$ on this set.

Since $X$ has $n$ elements, what can you say about the orbit of $g$ under this action?

N. S.
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