Let H be a normal subgroup of a group G, and let&nbsp;m=(G:H). Expose thatam∈Hfor every&nbsp;a∈G

Since H is a normal group of G and&nbsp;m=(G:H)&nbsp;then we have that the order of&nbsp;G/H&nbsp;is m. Hence for every a in G we gave that&nbsp;(aH)m=H, since the order of every element divides the order of the group,which suggests&nbsp;am∈HAnswerlogically follows from the assumption that&nbsp;|G/H|=m

Answered: "Let H be a normal subgroup of a..."

College
Algebra
Abstract algebra

Emeli Hagan

Answered question

2021-02-26

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

Answer & Explanation

okomgcae

Skilled2021-02-27Added 93 answers

Since H is a normal group of G and $m = (G : H)$ then we have that the order of $G / H$ is m. Hence for every a in G we gave that $(a H)^{m} = H$ , since the order of every element divides the order of the group,which suggests $a^{m} \in H$
Answer
logically follows from the assumption that $| G / H | = m$

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