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

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
asked 2021-02-26
Let H be a normal subgroup of a group G, and let m = (G : H). Show that
\(a^(m)inH\)
for every \(a in G\)

Answers (1)

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 \(|G/H|=m\)
0

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