Automorphism group of $C_{4} \times C_{2}$

Question

Let $C_{4} = \langle a \rangle$ a cyclic group of order 4, denoted multiplicatively. Let $C_{2} = \langle b \rangle$ a group of order 2, also denoted multiplicatively.

The automorphism group of $C_{4} \times C_{2}$ is dihedral of order 8. (It is exercise 1.5.14 in Robinson's Course, p. 30.) I have a proof, but it requires a lot of calculations.

Here is this proof. The elements of order 4 of $C_{4} \times C_{2}$ are

$(a, 1), (a, b), (a^{-1}, 1)$ and $(a^{-1}, b)$.

Among these elements, $(a, 1)$ and $(a, b)$ are distinct and not inverses of eachother. Thus,

(1) if $f$ is an automorphism of $C_{4} \times C_{2}$, then $f(a, 1)$ and $f(a, b)$ are two elements of order 4, distinct and not inverses of eachother.

On the other hand,

(2) $(a, 1)$ and $(a, b)$ generate $C_{4} \times C_{2}$.

Indeed, the inverse of $(a, 1)$ is $(a^{-1}, 1)$ and the inverse of $(a, b)$ is $(a^{-1}, b)$, thus the subgroup of $C_{4} \times C_{2}$ generated by $(a, 1)$ and $(a, b)$ contains the four elements of order 4 of $C_{4} \times C_{2}$. It also contains the identity element, thus it contains more than 4 elements, so it is the whole $C_{4} \times C_{2}$.

From (1) and (2), it results that $f \mapsto (f(a, 1), f(a, b))$ defines an injection from the set $Aut(C_{4} \times C_{2})$ into the set of odered pairs $(\alpha, \beta)$ such that $\alpha$ and $\beta$ are distinct elements of order 4 of $C_{4} \times C_{2}$ and are not inverses of eachother. There are exactly 8 such ordered pairs (4 possible values for $\alpha$ and, for each value of $\alpha$, two possible values for $\beta$). Thus

(3) $\vert Aut(C_{4} \times C_{2}) \vert \leq 8$.

Choose a numbering $\gamma_{1}, \ldots \gamma_{7}$ of the elements of $C_{4} \times C_{2}$ distinct from $(1, 1)$. If $f$ denotes a permutation of the set $C_{4} \times C_{2}$ such that $f(1,1) = (1, 1)$, then in order to verify that $f$ is an automorphism, it suffices to verify that $f(\gamma_{i}\gamma_{j}) = f(\gamma_{i}) f(\gamma_{j})$ for every $i$ and $j$ such that $1 \leq i \leq j \leq 7$ (since $C_{4} \times C_{2}$ is abelian).

This requires calculating 56 products (28 for the left member and 28 for the right member).

Now, let $f_2$ denote the permutation of $C_{4} \times C_{2}$ defined by $f_2(a, 1) = (a, 1), f_2(a, b) = (a^{-1}, b), f_2(a^{-1}, 1) = (a^{-1}, 1), f_2(a^{-1}, b) = (a, b), f_2(a^{2}, 1) = (a^{2}, 1), f_{2}(1, b) = (a^{2}, b), f_{2}(a^{2}, b) = (1, b), f_{2}(1, 1) = (1, 1).$

By calculating 56 products, we verify that $f_2$ is an automorphism of $C_{4} \times C_{2}$ and clearly this automorphism has order 2.

Let $f_4$ denote the permutation of $C_{4} \times C_{2}$ defined by $f_4(a, 1) = (a, b), f_4(a, b) = (a^{-1}, 1), f_4(a^{-1}, 1) = (a^{-1}, b), f_4(a^{-1}, b) = (a, 1), f_4(a^{2}, 1) = (a^{2}, 1), f_{4}(1, b) = (a^{2}, b), f_4(a^{2}, b) = (1, b), f_4(1, 1) = (1, 1).$

By calculating 56 products, we verify that $f_4$ is an automorphism of $C_{4} \times C_{2}$ and, clearly this automorphism has order 4. Moreover, $f_2 f_4 {f_2}^{-1} (= f_2 f_4 f_2) = {f_4}^{-1}$. A group generated by an element $x$ of order $n$ and an element $y$ of order 2 such that $y x y^{-1} (= y x y) = x^{-1}$ is dihedral of order $2n$, thus $Aut (C_{4} \times C_{2})$ contains a dihedral group of order 8. Thus, in view of (3), $Aut (C_{4} \times C_{2})$ is dihedral of order 8 and we have done.

The problem with this proof is that it requires calculating 112 products. So my question is : do you know a shorter elementary proof ? Thanks in advance.

Edit : there is a related question on this page : A question about the automorphism group of $\mathbb{Z}_{2} \times \mathbb{Z}_{4}$ but it seemes to me that this page doesn't answer my question completely.

Edit : it was pointed out that there is another proof on this page : Automorphisms and Mappings

but it seems to me that there too it is necessary to know explicitly some automorphisms, which needs at least 112 calculations. So my question remains.

Edit (July 7, 2023) No, there is no reply to my question on the page A question about the automorphism group of $\mathbb{Z}_{2} \times \mathbb{Z}_{4}$ . I ask for a proof at the level of Robinson's exercice, without presentations, without quaternion group and without classification of groups of order 8. If anybody says : "this is an automorphism of $C_{4} \times C_{8}$", please explain how it can be proved without calculating 56 products. That is my question.

Answer 1

You do not have to calculate all these products to check that each of your $f_k$'s ($1\le k\le8$) is a group morphism. You can define it as such.

For this, let us prove that for every ordered pair $(\alpha, \beta)$ such that $\alpha$ and $\beta$ are distinct elements of order 4 of $C_4\times C_2$ and are not inverses of eachother, there is a group endomorphism of $C_4\times C_2$ which maps $(a,1)$ to $\alpha$ and $(a,b)$ to $\beta.$ Note that $(a,1)^{-1}(a,b)=(1,b)$ and that $\alpha^{-1}\beta$ is of order $2.$

More generally, for every ordered pair $(\alpha,\gamma)$ of elements of $C_4\times C_2$ such that $\alpha^4=\gamma^2=(1,1),$ we shall construct the (unique) group endomorphism $f$ of $C_4\times C_2$ which maps $x:=(a,1)$ to $\alpha$ and $y:=(1,b)$ to $\gamma$.

Defining $f$ by: $$\forall (p,q)\in\{0,1,2,3\}\times\{0,1\}\quad f(x^py^q)=\alpha^p\gamma^q,$$ and using the hypothesis on $\alpha,\gamma,$ we immediately get: $\forall (p,q),(r,s)\in\{0,1,2,3\}\times\{0,1\}$ $$\begin{align}f(x^py^qx^ry^s)&=f(x^{p+r}y^{q+s})\\&=\alpha^{p+r}\gamma^{q+s}\\&=\alpha^p\gamma^q\alpha^r\gamma^s\\&=f(x^py^q)f(x^ry^s). \end{align}$$