# Semidirect Product variations

Is there two semidirect products of order $16$ with $C_2$ normal subgroup?

How many groups of the form $C_4 \times C_2$ : $C_2$ are there ?

Is it one expression for two groups ? or more?

• You are looking for maps $C_4 \times C_2 \to \mathrm{aut}(C_2)$. So ask yourself what $\mathrm{aut}(C_2)$ is. – Myself Aug 24 '14 at 21:26
• could you explain more ? what do you mean? – math math Aug 24 '14 at 21:27
• You should probably add to your question the definition of semi-direct product you're most comfortable with if you don't understand my comment. – Myself Aug 24 '14 at 21:28
• ok thanks for help – math math Aug 24 '14 at 21:29

A semi direct product $A\rtimes B$ is given by a group action of $B$ onto $A$, or equivalently by a group homomorphism $B\to \mathrm{Aut}(A)$, where $\mathrm{Aut}(A)$ is the group of automorphisms of $A$. Moreover, the semi direct product is isomorphic to $A\times B$ if the action is trivial.
In your case, the normal subgroup $A$ is of size $2$, so there are two elements, one is the identity and the other of order $2$. You cannot switch them, so $\mathrm{Aut}(A)$ is trivial. Hence, $A\rtimes B=A\times B$ for each group $B$, if $A$ is of size $2$.
If you try to do the converse, so take $A=C_4\times C_2$ and $B=C_2$, then you need to find an action of $B$ onto $A$.
There are two groups with structure description $(C_4\times C_2)\rtimes C_2$ whose Gap Ids are $(16,3)$ and $(16,13)$. You may see this for more detail.