# Prove every automorphism can be realized as conjugation

Let $N$ be a group, and $\alpha : N \to N$ be an automorphism. Prove that $\alpha$ may be realized as a conjugation. i.e., there exists a group $G$ containing $N$ as a normal subgroup with $\alpha(n) = gng^{-1}$ for some $g \in G$.

I'm guessing the way to do this is to construct $G$ as a semi-direct product of $N$ with some cleverly chosen group $H$...but I'm stuck at picking the right $G$ or $H$. This problem comes from Aluffi exercise IV.5.7

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You're very close. Remember that $\alpha$ is an element of the group $\operatorname{Aut}(N)$, so as ever there is a homomorphism $\theta\colon \mathbf{Z} \to \operatorname{Aut}(N)$ determined by $\theta(1) = \alpha$. This more or less does the job, although you could certainly say something extra in the case where $\theta$ has a non-trivial kernel.

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Great! Then the conjugation happens in $N \rtimes_{\theta} \mathbb{Z}$, and with some careful steps to not mix up the group operations in $\mathbb{Z}$ and $N$, the conjugation works out. –  JeremyKun Oct 15 '11 at 20:47
@Bean: Is your problem to realize each automorphism as conjugation, with possibly different automorphisms being realized as conjugations in different groups, or to find a single group in which every automorphism is realized as conjugation in that group? –  Arturo Magidin Oct 15 '11 at 21:08
The automorphism $\alpha$ is fixed from the beginning. Dylan's hint led me to the solution. –  JeremyKun Oct 15 '11 at 21:22
@Bean: Ah, I misunderstood the problem, then. The solution I give below gives you a group in which you can realize all automorphisms in the same group; Dylan's solution is far simpler if all you care about is a single automorphism. –  Arturo Magidin Oct 15 '11 at 21:43

The simplest thing is to remember that a semidirect product $N\rtimes K$ of groups $N$ and $K$ is equivalent to a group homomorphism $\phi\colon K\to\mathrm{Aut}(N)$: given a semidirect product $N\rtimes K$, we obtain $\phi$ by letting $\phi(k)$ be the automorphism induced on $N$ by conjugation by $k$ in $N\rtimes K$. Conversely, given a homomorphism $\phi\colon K\to\mathrm{Aut}(N)$, we construct the semidirect product $N\rtimes_{\phi}K$ as the group whose underlying set is the set $N\times K$ of all ordered pairs $(n,k)$ with $n\in N$, $k\in K$, and with group operation given by $$(n,k)\cdot (m,\ell) = (n\phi(k)(m), k\ell).$$ In this group, we have that $$(e,k)(n,e)(e,k)^{-1} = (e,k)(n,e)(e,k^{-1}) = (e\phi(k)(n)e,kk^{-1}) = (\phi(k)(n),e).$$ That is, conjugation by $(e,k)$ corresponds exactly to the application of $\phi(k)$ to $N$. We then identify $N$ with the subgroup $\{(n,e)\mid n\in N\}$, and $K$ with the subgroup $\{(e,k)\mid k\in K\}$.

So to get a semidirect product you want to find some group $K$ and a group homomorphism $\phi\colon K\to\mathrm{Aut}(N)$. Then conjugation by $k$ in $N\rtimes K$ will correspond to the automorphism $\phi(k)$. Since we want every automorphism to be represented, we want $\phi$ to be onto. And that's really all we need.

So... we just need to find some group $K$ that has an onto homomorphism to $\mathrm{Aut}(N)$. What's the simplest one and simplest $\phi$ we can find?

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