## Necessary and Sufficient Conditions for Semidirect Product to be Abelian (Proof)

This theorem is pretty basic, but it is useful to construct non-abelian groups. Basically, once you have either group to be non-abelian, or the homomorphism to be trivial, the end result is non-abelian!

Theorem: The semidirect product $N\rtimes_\varphi H$ is abelian iff $N$, $H$ are both abelian and $\varphi: H\to\text{Aut}(N)$ is trivial.

Proof:
$(\implies)$

Assume $N\rtimes_\varphi H$ is abelian. Then for any $n_1, n_2\in N$, $h_1, h_2\in H$, we have
\begin{aligned} (n_1, h_1)\cdot(n_2,h_2)&=(n_2,h_2)\cdot(n_1, h_1)\\ (n_1\varphi_{h_1}(n_2), h_1h_2)&=(n_2\varphi_{h_2}(n_1), h_2h_1). \end{aligned}
This implies $h_1h_2=h_2h_1$, thus $H$ is abelian.

Consider the case $n_1=n_1=n$. Then for any $n\in N$, $n\varphi_{h_1}(n)=n\varphi_{h_2}(n)$. Multiplying by $n^{-1}$ on the left gives $\varphi_{h_1}(n)=\varphi_{h_2}(n)$ for any $h_1, h_2\in H$. Thus $\varphi_h(n)=\varphi_e(n)=n$ for all $h\in H$ so $\varphi$ is trivial.

Consider the case where $h_1=h_2=e$. Then we have $n_1n_2=n_2n_1$, so $N$ has to be abelian.

($\impliedby$)

This direction is clear.

Advertisements

## About mathtuition88

http://mathtuition88.com
This entry was posted in math and tagged . Bookmark the permalink.