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Describe all semidirect products $Bbb Z/4 Bbb Z rtimes Bbb Z/2 Bbb Z$.

What the litterature says about this question: There is no such semidirect product because there is no non trivial morphisme from $Bbb Z/2 Bbb Z$ to $(Bbb Z/4 Bbb Z)^*$.

Is the simple fact $Bbb Z/4 Bbb Z cap Bbb Z/2 Bbb Z = 0,1 neq 0$ a good argument for the non existence of such product? Or is it wrong to say that $1 in Bbb Z/4 Bbb Z cap Bbb Z/2$ because $1$ has different orders in the same group?

Thank you for your help.

You write "there is no such semidirect product". That's not true, because we always have the trivial semidirect product, namely the direct product, that is $(mathbb Z / 4 mathbb Z) times (mathbb Z / 2 mathbb Z)$.

In order to search for more semidirect products, let's determine the automorphism group $A$ of $mathbb Z / 4 mathbb Z$.

$mathbb Z / 4 mathbb Z$ is the cyclic group of order 4. We can write its elements as 0, 1, 2, 3, where the group operation is addition mod 4.

From this, we see that $mathbb Z / 4 mathbb Z$ has two generators, namely 1 and 3.

Since $mathbb Z / 4 mathbb Z$ is generated by 1, any endomorphism $varphi: mathbb Z / 4 mathbb Z rightarrow mathbb Z / 4 mathbb Z$ is determined by its image $varphi(1)$ of 1.

And conversely, for any $g in mathbb Z / 4 mathbb Z$, the assignment $1 mapsto g$ defines an endomorphism $varphi: mathbb Z / 4 mathbb Z rightarrow mathbb Z / 4 mathbb Z$ with $varphi(1) = g$. This has to do with the fact that the order of $g$ is a divisor of 4 (the group order).

Also, it's "easy to see" that the endomorphism $varphi$ is an automorphism iff $varphi(1)$ is a generator of $mathbb Z / 4 mathbb Z$.

Since $mathbb Z / 4 mathbb Z$ has two generators (1 and 3, see above), we have two automorphisms, namely $varphi_1$ determined by $varphi_1(1) = 1$, and $varphi_3$ determined by $varphi_3(1) = 3$.

Now, it's "easy to see" that $varphi_1$ is the identity map, and "almost as easy to see" that $varphi_3$ has order 2, that is $varphi_3 circ varphi_3 = varphi_1$.

But this means that the automorphism group $A$ of $mathbb Z / 4 mathbb Z$ is the cyclic group of order 2, generated by $varphi_3$.

Now, because $A$ and $mathbb Z / 2 mathbb Z$ are so "simple" (each consists of only two elements, in fact they are isomorphic), there are only two homomorphisms from $mathbb Z / 2 mathbb Z$ into $A$. Let's write $mathbb Z / 2 mathbb Z = 0, 1 $ with addition mod 2 as group operation. Then $1$ is the generator of $mathbb Z / 2 mathbb Z$, and we have the homomorphisms $alpha_1: (mathbb Z / 2 mathbb Z) rightarrow A, 1 mapsto varphi_1$ and $alpha_3: (mathbb Z / 2 mathbb Z) rightarrow A, 1 mapsto varphi_3$.

$alpha_1$ is trivial and leads to the trivial semidirect product $(mathbb Z / 4 mathbb Z) times (mathbb Z / 2 mathbb Z)$.

$alpha_3$ is nontrivial and leads to the only other (necessarily nontrivial) semidirect product $(mathbb Z / 4 mathbb Z) rtimes (mathbb Z / 2 mathbb Z)$.

So we found that there are two semidirect products $(mathbb Z / 4 mathbb Z) rtimes (mathbb Z / 2 mathbb Z)$, one trivial and one nontrivial.

You also write $(mathbb Z / 4 mathbb Z) cap (mathbb Z / 2 mathbb Z) = 0, 1 $.
That's not correct, because without further context (for example both groups embedded into a larger group), $(mathbb Z / 4 mathbb Z)$ and $(mathbb Z / 2 mathbb Z)$ are different things and have empty intersection.

For example, it is true that, if we write $(mathbb Z / 4 mathbb Z) = 0, 1, 2, 3 $ as above, then the subset $0, 2 $ is isomorphic to $(mathbb Z / 2 mathbb Z)$ (check that), and thus we can write $(mathbb Z / 4 mathbb Z) cap (mathbb Z / 2 mathbb Z) = 0, 2 $.

But in general, if we write both groups as sets of residues mod 4 and 2, respectively, then we have
$$
(mathbb Z / 4 mathbb Z) = 0_4, 1_4, 2_4, 3_4
$$
and
$$
(mathbb Z / 2 mathbb Z) = 0_2, 1_2
$$
where I have written the modulus as subscript.
"Of course", $0_4 neq 0_2$ and $1_4 neq 1_2$, and thus
$$
(mathbb Z / 4 mathbb Z) cap (mathbb Z / 2 mathbb Z) = emptyset,
$$
and not
$$
(mathbb Z / 4 mathbb Z) cap (mathbb Z / 2 mathbb Z) = 0, 1 quad leftarrow WARNING: THAT'S WRONG.
$$