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Let $M$ be a subgroup of $G$, how do we get the property that if $x\in M$ then $xM=M$?

As far as I know the multiplication is only defined to be self-contained in $M$, how do I know that if I pick some $m\in M$ then there exists a $m'\in M$ such that $m=xm'$?

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  • $\begingroup$ Because both $m$ and $x^{-1}$ are in $M$. $\endgroup$ Feb 1, 2019 at 17:26
  • $\begingroup$ P.S. Please include your question in the body of the message, not just the title. The post should not depend on the title/subject for context. $\endgroup$ Feb 1, 2019 at 17:27
  • $\begingroup$ I can pick $m'=x^{-1}m$ $\endgroup$
    – RM777
    Feb 1, 2019 at 17:28

1 Answer 1

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Since $x $ and $m $ are in $M $, $x^{-1}m $ is also in $M $. Take $m'=x^{-1}m$.

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