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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.

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  • $\begingroup$ You're right, I've edited my post right away. Dumb mistake. $\endgroup$ – Matheus Andrade May 11 '18 at 20:09
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    $\begingroup$ No. $K$ can be trivial, and $m$ can be element of order $2$. $\endgroup$ – SMM May 11 '18 at 20:09
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    $\begingroup$ No. In the integers modulo $2$ we have $1+1 = 0$ in the subgroup $K$ consisting only of $0$. $\endgroup$ – Ethan Bolker May 11 '18 at 20:10
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No way. In the group $\mathbb{Z}$ consider the subgroup $E$ of evens. Then $1+1 \in E$ but $1 \notin E$.

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  • $\begingroup$ 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. $\endgroup$ – Matheus Andrade May 11 '18 at 20:10
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No: Let $K=\mathbb{Q}^*$ and $G = \mathbb{R}^*$. Then $\sqrt{2}*\sqrt{2}\in K$ but $\sqrt{2}\notin K$.

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    $\begingroup$ The reals aren't a group under multiplication. $\endgroup$ – Randall May 11 '18 at 20:13
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    $\begingroup$ Now we're talking. Good example. $\endgroup$ – Randall May 11 '18 at 20:14
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Hint $G=\mathbb Z, K=2 \mathbb Z$.

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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.

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