# Prove there's only one nonabelian semidirect product $(C_2 \times C_2) \ltimes C_3$ up to isomorphism

I am trying to prove that there is only one nonabelian semidirect product $$(C_2 \times C_2) \ltimes C_3$$ up to isomorphism. By the definition of semidirect product, I am trying to find a group homomorphism $$\phi : C_2 \times C_2 \rightarrow \text{Aut}(C_3)$$. Initially, I wanted to prove this via cases. Note that $$\text{Aut}(C_3) = \{\text{id}, (1 \ 2)\}$$, where $$(1 \ 2)$$ is the permutation that switches $$1$$ and $$2$$ and id is the identity mapping. Since the trivial case produces an abelian group, the problem reduces to showing the following identifications of $$\phi$$ all yield isomorphic groups $$(C_2 \times C_2) \ltimes C_3$$ (as each produces a nonabelian semidirect product): (1) $$\phi(0,1) = (1 \ 2)$$, $$\phi(1,0) = \text{id}$$, and thus $$\phi(1,1) = (1 \ 2)$$; (2) $$\phi(0,1) = \text{id}$$, $$\phi(1,0) = (1 \ 2)$$, and thus $$\phi(1,1) = (1 \ 2)$$; and (3) $$\phi(0,1) = (1 \ 2)$$, $$\phi(1,0) = (1 \ 2)$$, and thus $$\phi(1,1) = \text{id}$$. However, this has become too tedious for my liking, and I'm becoming muddled in finding isomorphisms between each case. Is there a cleverer way of showing this?

## 1 Answer

You could argue that all those morphisms $$C_2\times C_2\to \operatorname{Aut}(C_3)$$ are conjugated, in the sense that for any two of them (say $$f$$ and $$g$$), there is an automorphism $$\phi$$ of $$C_2\times C_2$$ such that $$f=g\circ \phi$$.

This is easy to see since all those morphisms are basically the non-zero linear maps $$\mathbb{F}_2^2\to \mathbb{F}_2$$, so this is just basic linear algebra.

I think you can then easily establish that if those morphisms are conjugated, they induce isomorphic semi-direct products (this is true in general, for any groups).

• Did you mean "conjugate," rather than conjugated? – Derek Adams Nov 1 '19 at 0:46