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If $H \lhd G$ and $[G:H] = n,$ then $a^n \in H$ for all $a ∈ G.$

I know that $H$ is a normal subgroup of $G$ with the index of $G:H = n.$ This means that $|G|/|H|=n,$ so $|G|=|H|n. $

I attempted to say that $a^{|G|} = a^{|H|}n = e,$ the identity, but have gotten stuck here, and have also tried to interpret the question in the quotient group $G/H.$

If someone could provide a thorough proof I'd appreciate it.

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    $\begingroup$ If $H\triangleleft G$, $G/H$ is a group with cardinality $[G:H]$. Hence by Lagrange's theorem... $\endgroup$ – Maxime Ramzi Apr 25 '18 at 21:40
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There is a stronger statement:

$$n=[G:H]=|G/H|$$ In other words, there is a homomorphism $\phi:G \to G/H$, and for any $\phi(a) \in G/H$, we have that $\phi(a^n)=\phi(a)^n=1$ which occurs if and only if $a^n \in \ker \phi$ or, $a^n \in H$.

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  • $\begingroup$ How can we prove $a^n∈kerϕ$? $\endgroup$ – BOA2018 Apr 25 '18 at 21:57
  • $\begingroup$ $\phi(a)^n=1$ since $n$ is the size of the group, so $\langle \phi(a) \rangle $ divides $|G/H|$ by lagranges, theorem. So $a$ is in the kernel by definition... $\endgroup$ – Andres Mejia Apr 25 '18 at 21:58

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