H is a subgroup of a finite group G, and if |G| = m|H|, adapt the proof of Lagrange's Theorem to show that g^m! ∈ H for all g ∈ G.

Question

I'm finding this question particularly difficult to answer. Could someone help me, please?

Edit: I have seen the use of the notation ker which I know translates to kernel, in some solutions, but this notation is not used in my course on Group Theory. What is the alternative solution to this?

Armstrong question 11.11

See also https://math.stackexchange.com/questions/545417/if-gh-n-is-it-true-that-xn-in-h-for-all-x-in-g — lhf, Jun 02 '17 at 12:07
But I wonder what the book intends by suggesting that we "adapt the proof of Lagrange's Theorem". — lhf, Jun 02 '17 at 13:16

score 1 · Accepted Answer · answered Jun 02 '17 at 12:04

1

Let $X$ be the set of left cosets of $H$. Then $\phi: G \to \text{Sym}(X)$ given by $\phi(g)(aH)=(ga)H$ is a homomorphism. Therefore, by Lagrange's Theorem, $id = \phi(g)^{m!}= \phi(g^{m!})$ for all $g \in G$. In particular, $H=\phi(g^{m!})(H)=g^{m!}H$ and so $g^{m!} \in H$.

answered Jun 02 '17 at 12:04

lhf

216,483

I'm not sure this answer fits the book's suggestion that we "adapt the proof of Lagrange's Theorem". – lhf Jun 02 '17 at 13:20
Yes, that must be what is meant by that condition. – anon Jun 03 '17 at 00:48
I know this is a novice question to ask, but what is the definition of the notation Sym? That notation has never appeared in my textbook, so I don't really understand what is going on here. – wsh_97 Jun 03 '17 at 00:48
@Goooie, $\text{Sym}(X)$ is the group of bijections (permutations) of $X$. – lhf Jun 03 '17 at 01:59

H is a subgroup of a finite group G, and if |G| = m|H|, adapt the proof of Lagrange's Theorem to show that g^m! ∈ H for all g ∈ G.

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