# Let $G$ be a group and $m \in G$, $K$ be a subgroup and $m * m \in K$. Is it always true that $m \in K$?

I'm trying to use this in another proof, but I think it might be false and I'm out of ideas to prove it.

• You're right, I've edited my post right away. Dumb mistake. – Matheus Andrade May 11 '18 at 20:09
• No. $K$ can be trivial, and $m$ can be element of order $2$. – SMM May 11 '18 at 20:09
• No. In the integers modulo $2$ we have $1+1 = 0$ in the subgroup $K$ consisting only of $0$. – Ethan Bolker May 11 '18 at 20:10

No way. In the group $\mathbb{Z}$ consider the subgroup $E$ of evens. Then $1+1 \in E$ but $1 \notin E$.

• You're right, I'm glad I didn't try to use this. I'll accept your answer as soon as possible. Very silly mistake on my part. – Matheus Andrade May 11 '18 at 20:10

No: Let $K=\mathbb{Q}^*$ and $G = \mathbb{R}^*$. Then $\sqrt{2}*\sqrt{2}\in K$ but $\sqrt{2}\notin K$.

• The reals aren't a group under multiplication. – Randall May 11 '18 at 20:13
• Now we're talking. Good example. – Randall May 11 '18 at 20:14

Hint $G=\mathbb Z, K=2 \mathbb Z$.

Not necessarily.

Consider the rationals as a subgroup of real numbers.

$\sqrt 5 \times \sqrt 5$ is in the subgroup but $\sqrt 5$ is not.