# Classifying groups of order 60

I want to solve the following problem from Dummit & Foote's Abstract Algebra text (p. 185, Exercise 14):

This exercise classifies the groups of order $60$ (there are thirteen isomorphism types). Let $G$ be a group of order 60, let $P$ be a Sylow 5-subgroup of $G$ and let $Q$ be a Sylow 3-subgroup of $G$.

(a) Prove that if $P$ is not normal in $G$ then $G \cong A_5$. [See Section 4.5.]

(b) Prove that if $P \trianglelefteq G$ but $Q$ is not normal in $G$ then $G \cong A_4 \times Z_5$. [Show in this case that $P \leq Z(G)$,$G/P \cong A_4$, a Sylow 2-subgroup $T$ of $G$ is normal and $TQ \cong A_4$.]

(c) Prove that if both $P$ and $Q$ are normal in $G$ then $G \cong Z_{15} \rtimes T$ where $T \cong Z_4$ or $Z_2 \times Z_2$. Show in this case that there are six isomorphism types when $T$ is cyclic (one abelian) and there are five isomorphism types when $T$ is the Klein 4-group (one abelian). [Use the same ideas as in the classifications of groups of orders 30 and 20.]

My attempt:

(a) If $n_5(G)>1$ then $G$ is simple by Proposition 21 of Section 4. Proposition 23 of the same section gives $G \cong A_5$.

(b) If $P \trianglelefteq G$ but $Q \not \trianglelefteq G$, Sylow's Theorem part (iii) gives $n_3(G) \geq 4$. Let $Q_1,Q_2,Q_3,Q_4$ be four distinct Sylow 3-subgroups. The groups $PQ_i$ are all cyclic and intersect (pairwise) at $P$ (since they have $2$ elements of order $3$). Thus $C_G(P)$ contains at least $15+10+10+10=45$ elements. Lagrange's Theorem gives $|C_G(P)|=60$ and consequently $Z(G) \geq P$.

The quotient $G/P$ is of order $12=2^2 \cdot 3$. We know that for groups of order 12 either $n_2=1$ or $n_3=1$. Suppose by way of contradiction that $n_3=1$. As the group $Q_1P/P$ is a Sylow 3-subgroup it is normal. The Fourth Isomorphism Theorem gives that $Q_1P \trianglelefteq G$. However, let $g$ be an element of $G$ conjugating $Q_1$ to $Q_2$ (which exists by Sylow's Theorem part (ii)). We have $gQ_1Pg^{-1}=gQ_1g^{-1}gPg^{-1}=Q_2P \neq Q_1P$. Thus $n_3 \neq 1$ and consequently $n_2=1$. The discussion on pages 182-183 gives $G/P \cong A_4$. The Fourth Isomorphism Theorem applied again gives that a Sylow 2-subgroup of $G$, $T$ is normal. Thus $TQ \leq G$ is a subgroup of order $12$. If $Q \trianglelefteq TQ$ then since $Q$ is normalised by $TQ$ as well as the elements of the central subgroup $P$, we find that $Q \trianglelefteq G$ which is a contradiction. Thus $n_3(TQ) \neq 1$ and consequently $TQ \cong A_4$. Now, since $P \trianglelefteq G$ and $P \cap TQ=1$, we have that $G \cong P \rtimes_\varphi TQ \cong Z_5 \rtimes_\varphi A_4$ for some homomorphism $\varphi$. Since $A_4$ is generated by the 3-cycles, and since $\text{Aut}(Z_5) \cong Z_4$ has no elements of order 3, we find that $\varphi$ is trivial. Hence $G \cong Z_5 \times A_4$ in that case.

(c) If $P,Q \trianglelefteq G$, then $PQ \trianglelefteq G$ is a normal subgroup, with complement $T$, a Sylow 2-subgroup of $G$. $PQ$ is a group of order $15$, so it must be isomorphic to $Z_{5} \times Z_3=\langle a \rangle \times \langle b \rangle$, while $T$ has two possible isomorphism types: $Z_4$ and $Z_2 \times Z_2$. The semidirect product criterion gives $G \cong (Z_{5} \times Z_3) \rtimes_\varphi T$ for some homomorphism $\varphi:T \to \text{Aut}(Z_{5} \times Z_3) \cong Z_4 \times Z_2$.

We begin with the case where $T =\langle x \rangle \cong Z_4$ is cyclic. $\varphi$ is then determined by $\varphi(x)$ which can have orders $1,2$ or $4$. If $|\varphi(x)|=1$ we get the direct product $Z_{15} \times Z_4$. If $|\varphi(x)|=2$, since we have 3 elements of order 2 in $\text{Aut}(PQ)$ as listed on page 182. If $\varphi(x)$ sends $a \mapsto a$ and $b \mapsto b^{-1}$ we get the following presentation

\begin{equation} G_1=\langle a,b,x| a^5=b^3=x^4=1|bab^{-1}=a,xax^{-1}=a,xbx^{-1}=b^{-1} \rangle \end{equation}

which can be factored as $G_1 \cong \langle a \rangle \times \langle b,x \rangle$. Since $\langle b,x \rangle$ is a group of order $12$ which has $n_3=1$, it is isomorphic to the one of example 2 on page 178.

This is pretty much where I get stuck... I can get presentations for the other semidirect products, but I can't see which ones are isomorphic to each other. I am aware of the solution here, yet I think it uses too many unnatural lemmas.

My questions are:

1. Are my solutions to parts (a) and (b) correct?
2. Can you please help me solve part (c)? Simple solutions are preferred (that is, ones using tools from earlier in the text.)

Thank you!

• I don't like where you used presentations and relations, IMHO that's more advanced and it's not treated properly in Dummit and Foote's book. I also remember giving up on this exercise. – Zero Jan 27 '15 at 2:42
• I can't understand how the Correspondence Theorem gives rise to a normal Sylow 2-subgroup $T$, of $G$. – Alan Wang Feb 20 '17 at 9:55
• @AlanWang I suppose it goes something like this: If there exist two distinct Sylow 2-subgroups $T_1 \neq T_2$ in $G$, then $T_1 P$ and $T_2 P$ give two distinct subgroups of $G$ of order 20. They correspond to two distinct subgroups (this is where the fourth isomorphism theorem comes in) of $G/P$, of order $4$. The fact that $A_4$ has a unique subgroup of order $4$ provides a contradiction. – user1337 Feb 20 '17 at 11:09

To define a semidirect product $Z_{15} \rtimes T$, for $T = Z_4, Z_2 \times Z_2$, you must pick a morphism $T \to \operatorname{Aut} Z_{15}$, which will represent the conjugation action of $T$ on $Z_{15}$.
I think the key here is to recognize that this morphism is characterized by the group structure of $G$, and is independent of the copy of $T$ that you've selected within $G$. The reason is that conjugation defines a morphism $G \to \operatorname{Aut} Z_{15}$. Since $Z_{15}$ is Abelian, this map factors through $G/Z_{15}$, so the conjugation action of $T$ is $T \cong G/Z_{15} \to \operatorname{Aut} Z_{15}$.
Therefore the problem of classifying the isomorphism types of the semidirect products is the same as that of classifying the "types" of morphisms $\varphi \colon T \to \operatorname{Aut} Z_{15}$, where two morphisms $\varphi_1$ and $\varphi_2$ are considered equivalent if there exists an automorphism $\alpha$ of $T$ such that $\varphi_2 = \varphi_1 \circ \alpha$.
In practice, this means enumerating the subgroups $H$ of $\operatorname{Aut} Z_{15} \cong Z_2 \times Z_4$ which could possibly be a homomorphic image of $T$, and then determining the "abstract" ways that $H$ can be obtained as a quotient of $T$.
In the present situation, there is only ever one abstract way to obtain $\{1\}$, $Z_2$, $Z_4$ (or $Z_2 \times Z_2$, depending on $T$) as a quotient of $T$, so the problem really amounts to enumerating the subgroups of order $1$, $2$ and $4$ of $\operatorname{Aut} Z_{15}$. There are five subgroups that can be a quotient of $Z_2 \times Z_2$, and six that can be a quotient of $Z_4$.