I am trying to use a semi-direct product to construct a non-abelian group of order 75 (Ex 5.5.8 in Dummit & Foote). Using the third Sylow theorem, we get $n_5=1$ so the subgroup of order $25$ is normal. Hence it makes sense to use the following semi-direct product:

$$ (\text{Sylow }5\text{-subgroup})\rtimes(\text{Sylow }3\text{-subgroup}) $$

There are two groups of order $25$, either $\mathbb{Z}/5\mathbb{Z}\times \mathbb{Z}/5\mathbb{Z}$ or $\mathbb{Z}/25\mathbb{Z}$. In order to use the semi-direct product, I need a homomorphism from $\mathbb{Z}/3\mathbb{Z}$ to one of these. The automorphism group of $\mathbb{Z}/25\mathbb{Z}$ is $C_{20}$ so we can't use this group (since $3\nmid 20$). My questions are:

$$ \begin{split}&1&) \text{ What is } \mathrm{Aut}\left(\mathbb{Z}/5\mathbb{Z}\times \mathbb{Z}/5\mathbb{Z}\right)? \\ &2&) \text{ How can we construct a non-trivial homomorphism } \phi:\mathbb{Z}/3\mathbb{Z}\to \mathrm{Aut}(\mathbb{Z}/5\mathbb{Z}\times \mathbb{Z}/5\mathbb{Z}) \end{split} $$


1 Answer 1


$\mathbb{Z}_5\times \mathbb{Z}_5$ can be considered as a vector space over field $\mathbb{Z}_5$ (the addition is same as addition in abelian group modulo $5$, and scalar multiplication actually comes from addition: $v+v=2.v$, $v+v+v=3.v$, $\cdots$)

Thus any automorphism of the group $\mathbb{Z}_5\times \mathbb{Z}_5$ is also an automorphism of the vector space $\mathbb{Z}_5\times \mathbb{Z}_5$ and conversely (because of remark on scalar multiplication made earlier).

It is well known, what is the automorphism group of $\mathbb{Z}_5\times \mathbb{Z}_5$? It is $$ GL(2,5)= \begin{Bmatrix} \begin{bmatrix} a & b\\ c & d \end{bmatrix}\colon a,b,c,d\in\mathbb{Z}_5, ad-bc\neq 0 \end{Bmatrix}. $$ To get an element of order $3$ in $Aut(\mathbb{Z}_5\times \mathbb{Z}_5)$, we have to find a non-identity matrix $A$ such that $A^3=I$. Thus, $A$ satisfies polynomial $x^3-1=(x-1)(x^2+x+1)$. The quadratic factor has no root in $\mathbb{Z}_5$ (check) so it is irreducible. Further, $A\neq I$, hence the minimal polynomial of $A$ should divide quadratic factor $x^2+x+1$. But the minimal polynomial of $A$ has degree $\leq 2$ (size of $A$), so $x^2+x+1$ is the minimal (hence characteristic polynomial of $A$). Can we find a matrix $A$ explicitly with such characteristic polynomial? Yes; companion matrix $$ A= \begin{bmatrix} 0 & -1\\ 1 & -1 \end{bmatrix}. $$ Now we come to construction of group. Let $\mathbb{Z}_5\times \mathbb{Z}_5=\langle x,y\rangle$ with $x=\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $y=\begin{bmatrix} 0 \\ 1 \end{bmatrix}$. Then $Ax=y$ and $Ay=-x-y$ (check).

Consider $\mathbb{Z}_5\times \mathbb{Z}_5=\langle x,y\rangle$ multiplicatively, $t$ be element of order $3$ in its automorphism group, and action of $t$ on this group is given by (write above action of $A$ as action of $t$, with multiplicative operation): $$txt^{-1}=y \,\,\,\, tyt^{-1}=x^{-1}y^{-1}.$$ Thus $$G=\langle x,y,t\colon x^5=y^5=1, xy=yx, t^3=1, txt^{-1}=y, tyt^{-1}=x^{-1}y^{-1}\rangle$$ is a group of order $75$, it is non-abelian.

  • $\begingroup$ Thanks @Groups! I didn't think of viewing $\mathbb{Z}_{5}\times \mathbb{Z}_5$ as a vector space but this makes things much clearer and gives a concrete action for the semi-direct product. $\endgroup$ Nov 16, 2015 at 21:36

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