0

Suppose $G$ is a finite abelian group and $H$ be a proper subgroup. Let $a$ be an element in $G$ not in $H$. Does there always exists an $m>0$ s.t. $a^m \in H$? If it is there what is the proof?

John P.
  • 2,136

2 Answers2

0

This holds for any group $G$ and subgroup $H$ where $a \notin H$ has finite order: choose $m$ to be the order of $a$ so that $a^m = e \in H$.

angryavian
  • 89,882
0

More generally, this holds for any group $G$ where $H$ is a normal subgroup of finite index. This is because in the quotient group $G/H$ we have that the image of $a$ has finite order, say $k$. This means that $a^k\in H$.

Matt Samuel
  • 58,164