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.

  • 5
    $\begingroup$ Is "non-split extension" what you are looking for? $\endgroup$ Aug 5, 2023 at 8:00
  • $\begingroup$ yea you got it! $\endgroup$
    – Leon Kim
    Aug 5, 2023 at 8:17
  • 3
    $\begingroup$ 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". $\endgroup$ Aug 5, 2023 at 8:22

1 Answer 1


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).


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