T or F. If $G$ is an abelian group then G has no normal subgroups.

Asked Nov 05 '16 at 17:18

Active Nov 05 '16 at 17:18

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My attempt: False. Counterexample:

Let $G=\mathbb{R}$ and let $H$ be a subgroup of $G$ such that $H=2\mathbb{R}$.

Let $g=2, h=4$.

Then $ghg^{-1}=(2)(4)(1/2)=4$ and $4\in H$. Thus, $H$ is a normal subgroup of $G$.

My question: Is my counter-example correct? when i say "let $g=2$ and $h=4$" should $g$ and $h$ be an element of $G$ or $H$?

asked Nov 05 '16 at 17:18

combo student

2

The trivial subgroup is normal in all groups. – Adam Hughes Nov 05 '16 at 17:19
3

$G$ itself is always a normal subgroup. If $G$ is abelian, then any subgroup is normal, see here. So your example should work. $g=2$ should be element of $G=(\mathbb{R},+)$. – Dietrich Burde Nov 05 '16 at 17:20
shouldn't it be $G= (\mathbb{R}, *)$ in my case? @DietrichBurde – combo student Nov 05 '16 at 17:25
$(\mathbb{R},\cdot)$ is not a group! – Dietrich Burde Nov 05 '16 at 17:26
oh that's right! Thank you @DietrichBurde. So, $G=(\mathbb{R},+)$ and $H$ (subgroup of $G$) = $(2\mathbb{R},+)$. Let $g=2$ and $h=4$ such that $g,h\in G$. Then $ghg^{-1}=2+4+(-2)=4$. and $4$ exists in $H$. Is that correct now? – combo student Nov 05 '16 at 17:32
Correct. The next step for you now should be to review the definition of normal subgroup in the case that $G$ is abelian, not with some example, but in general. – Dietrich Burde Nov 05 '16 at 19:06

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