Comments on HW set 3
Problem 2
We want to show that there is exactly one group $G$ of order $75$ up to isomorphism.
If $G$ has order $75$, then it has a Sylow $5$-subgroup $H$ of order $25$ and a Sylow $3$ subgroup $K$ of order $3$. Notice that $H\cap K$ must be trivial since $H$ and $K$ have coprime orders.
The number $n_5$ of Sylow $5$-subgroups must both divide $3$ and also be congruent to one mod $5$, so $n_5=1$. This means $H$ is normal. Since all groups of order $p^2$, where $p$ is prime, are abelian, $H$ must be abelian. Therefore $H$ is either $\Zn{25}$ or $\Zn{5}\times \Zn{5}$.
Meanwhile, the number $n_3$ of Sylow $3$-subgroups must divide $25$ and be congruent to $1$ mod $3$. Therefore either $n_3=1$ or $n_3=25$. If $n_3=1$, then $K$ is normal. In this case, $HK$ is abelian and so $G$ is either $\Zn{5}\times\Zn{15}$ or $\Zn{75}$.
If $n_3=25$, then $K$ is one of $25$ conjugate subgroups of $G$ of order $3$ and $G=HK$ is the semidirect product of $H$ with $K$ arising from a map $K\to \Aut(H)$.
If $H=\Zn{25}$, then $\Aut(H)$ is of order $\phi(25)=20$, and there are no non-trivial maps from $K$ to $\Aut(H)$, so in this case the only group is the abelian group $\Zn{75}$.
If $H=\Zn{5}\times\Zn{5}$, then $\Aut(H)$ is $\GL_{2}(\Zn{5})$ which has order $480$. By Cauchy’s theorem, $\Aut(H)$ has an element $\sigma$ of order $3$. We can fix an isomorphism of $K$ with $\Zn{3}$ and define a map $\Zn{3}\to\Aut(H)$ sending $1$ to $\sigma$. The resulting semidirect product is nonabelian.
Now we need to prove that this semi-direct product is unique up to isomorphism. We can use problem 1 for this, but it is a little subtle.
Looking at the construction of the semi-direct product, we needed to choose an element $\sigma$ of order $3$ in $\GL_{2}(\Zn{5})$ and then we defined \(\phi: \Zn{3}\to \GL_{2}(\Zn{5})\) by sending $\phi(1)=\sigma$. The element $\sigma$ generates a Sylow 3-subgroup in $\GL_{2}(\Zn{5})$; call this $A(\sigma)\subset\GL_{2}(\Zn{5})$.
The first thing to notice is that, by Sylow’s theorem (which is stronger than Cauchy’s Theorem), every Sylow 3-subgroup of is of order 3 and they are all conjugate. This is because $480$ is divisible by $3$ but not $9$, so these elements of order $3$ actually generate Sylow 3-subgroups, all of which are conjugate.
Suppose we had chosen a different element of order $3$ in our construction, say $\sigma’$. Then the subgroup generated by $\sigma’$ would be conjugate to $A(\sigma)$. This means we can find an automorphism $g$ in $\GL_{2}(\Zn{5})$ that conjugates $A(\sigma’)$ into $A(\sigma)$.
In other words, $gA(\sigma’)g^{-2}=A(\sigma)$. However, it could happen that when we do this conjugation, that $g\sigma’g^{-1}=\sigma^{-1}$ because both $\sigma$ and $\sigma^{-1}$ generate $A(\sigma)$. However, we have an automorphism $f: A(\sigma)\to A(\sigma)$ such that $f(\sigma^{-1})=\sigma$.
In other words, any two nontrivial maps $\phi, \tau:\Zn{3}\to \Aut(\Zn{5}\times\Zn{5})$ can be changed into each other by finding $g\in \Aut(\Zn{5}\times \Zn{5})$ and $f\in \Aut(\Zn{3})$ so that \(\phi = \gamma_{g}\circ \tau \circ f\) and therefore by problem 1 the corresponding groups are isomorphic.