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The question is:

Let $G$ be a group and N $\unlhd G$ with $N$ not contained in $Z(G)$.

Prove that:

a) if $N \cong \bf{C}_3$, then $G$ has a subgroup of index $2$.
b) if $N \cong \mathbb{Z}$, then $G$ has a subgroup of index $2$.

How would I go about proving this? Would I have to explicitely construct the subgroups, or is there some theorem or indirect existence argument I can use?

Edit: Do I somehow have to construct two cosets in G?

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a) is false for $G=N=C_3$. –  Martin Brandenburg Oct 24 '12 at 16:21
    
My bad, I dropped a "not" in the "not contained in the center" part. –  Sirzh Oct 24 '12 at 16:24
1  
The equality $g^{-1}ng = n^2$ does not hold for all $g$ (for example it does hold if $g \in Z(G)$), but for each $n \in N$, there exists $g \in G$ such that we have $g^{-1}ng = n^2$ (take any $g$ that does not commute with $n$). –  Joel Cohen Oct 24 '12 at 16:32

1 Answer 1

up vote 3 down vote accepted

Since $N$ is normal, $G$ operates on $N$ by conjugation. Since $N$ is not in the center, this operation is not trivial, i.e. there is a nontrivial homomorphim $\phi\colon G\to\operatorname{Aut}(N)$. For both cases a) and b), we have $\operatorname{Aut}(N)\cong C_2$, so that the kernel of $\phi$ has index 2 in $G$.

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I'm sorry, but could you expand on why there's a nontrivial homomorphism? I'm not sure I get why. –  Sirzh Oct 24 '12 at 16:42
    
@Sirzh If $N$ is not in the center, it has some element $n\in N$ which doesn't commute with some other element $y_0\in G$, so $n^{y_0}\not= n$. Thus the homomorphism $\phi:G\rightarrow \text{Aut}(N)$ given by $\phi(g)=\sigma_g$, where $\sigma_g(x)=x^g$, is not the trivial, since $\sigma_{y_0}$ is not the identity. –  Alexander Gruber Oct 25 '12 at 7:27

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