# 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. [duplicate]

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

• Hint: Consider the cosets $H$, $Hg$, $Hg^2$, ... – Derek Holt Jun 2 '17 at 11:32
• – lhf Jun 2 '17 at 12:07
• But I wonder what the book intends by suggesting that we "adapt the proof of Lagrange's Theorem". – lhf Jun 2 '17 at 13:16

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 $g^{m!} \in H$.
• @Goooie, $\text{Sym}(X)$ is the group of bijections (permutations) of $X$. – lhf Jun 3 '17 at 1:59