# Is there a type of product of groups where $C_2 \star C_2 = C_4$?

Hi I am wondering if there is a type of product $$\star$$ of groups to get $$C_4 \cong C_2 \star C_2$$.

We know that the composition series of $$C_4$$ indeed is a 2 copy of $$C_2$$ so I am wondering if there is a notion of product to recover it back.

• Is "non-split extension" what you are looking for? Aug 5, 2023 at 8:00
• yea you got it! Aug 5, 2023 at 8:17
• The "usual" products are $C_2\times C_2$, or $C_2\ltimes C_2$, or the free product $C_2\ast C_2$. None of them is $C_4$. Of course, we have the nonsplits extension $1\rightarrow C_2\rightarrow C_4\rightarrow C_2\rightarrow 1$, but this seems a bit "constructed". Aug 5, 2023 at 8:22

As the comments pointed out, you can certainly call $$1 \to C_2 \to C_4 \to C_2 \to 1$$ a "non-split extension," but I don't think this is a 'product' in the sense that the equation $$C_2 \star C_2 = C_4$$ might suggest. I'll answer a question closer in spirit to the title: Given two groups $$A, G$$ and maybe some additional data, $$d$$, is there a way to construct a new group $$A \star_d G$$ that fits into the short exact sequence $$1 \to A \to A \star_d G \to G \to 1$$ such that if $$A = C_2 = G$$, there is some $$d$$ such that $$C_2 \star_d C_2 = C_4$$?
In fact there is! (or at least when $$A$$ is abelian) Under the hood, there is some moderately complicated math (group cohomology) but it isn't too hard to define $$A \star_d G$$ by just writing down the group operation.
We need two pieces of additional data: 1. a group action of $$G$$ on $$A$$ and 2. a cohomology class $$[f] \in H^2(G , A)$$. The only thing you need to know about group cohomology for the moment is that a second cohomology class $$[f]$$ can be represented by a function $$f:G^2 \to A$$.
With this data, we can define a group operation on the set $$A \times G$$ by $$(a,g) \cdot (b, h) = (a + g\cdot b + f(g,h), gh).$$ For the case $$A = C_2 = G$$, the trivial action of $$G$$ on $$A$$, and the cohomology class given by $$f(0,0) = f(1,0) = f(0,1) = 1$$ and $$f(1,1) = 0$$, we have $$(1,1)(1,1) = (0,0)$$ $$(0,0)(1,1) = (0,1)$$ $$(0,1)(1,1) = (1,0)$$ $$(1,0)(1,1) = (1,1)$$ which you can verify is cyclic of order 4, generated by $$(1,1)$$, with trivial element $$(1,0)$$. You can also check that the trivial cohomology class given by $$f(x,y) = 0$$ gives rise to the direct product $$A \times G$$.
In fact, there is a bijective correspondence between $$H^2(G,A)$$ and extensions with the given group action. If you feel like nuking a mosquito, you can use this fact to classify all groups of order 4 by calculating $$H^2(G,A)$$ (one of my favorite exercises).