# Let H be a normal subgroup of a group G, and let m = (G : H). Show thata^(m)inHfor every a in G

Let H be a normal subgroup of a group G, and let $$m = (G : H)$$. Show that
$$a^{m} \in H$$
for every $$a \in G$$

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okomgcae

Since H is a normal group of G and $$m=(G:H)$$ then we have that the order of $$G/H$$ is m. Therefore for every a in G we gave that $$(aH)^m =H$$, since the order of every element divides the order of the group,which implies $$a^m \in H$$
Result
Follows from the fact the assumptions imply $$|G/H|=m$$