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.


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.


This direction is clear.

Author: mathtuition88

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