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

Question
Abstract algebra
Let H be a normal subgroup of a group G, and let m = (G : H). Show that
$$a^(m)inH$$
for every $$a in G$$

2021-02-27
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 $$a^m inH$$
Result
Follows from the fact the assumptions imply $$|G/H|=m$$

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