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:


*

*Are my solutions to parts (a) and (b) correct?

*Can you please help me solve part (c)? Simple solutions are preferred (that is, ones using tools from earlier in the text.)


Thank you!
 A: My answer only concerns part (c).
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 $Z_4$.
A: Studying this problem too.
Take shorthand for cyclic group $Z_n := \mathbb{Z} /n\mathbb{Z}$.
All references come from Dummit & Foote, Abstract Algebra, 3e.
You are doing well, I just simplify and clarify some parts of your proof.
For part (b), you showed $P \in Z(G)$, and either $G/P \cong A_4$ or $S_3 \times Z_2$ (D&F, p.182).
If $G/P \cong A_4$, there must be $G \cong A_4 \rtimes P$ (D&F p.180), and a moment of reflection shows the product is trivial by $P \in Z(G)$, that is, $G \cong A_4 \times P$.
It remains to rule out $G \cong S_3 \times Z_2$.
To reach contradiction, I shall show that the Sylow-3 $G_3 \lhd G$.
By isomorphism thm. 4, find the preimage of $S_3$ in $G/P$, call it $R$, and $|R|=30$.
We saw that there is a order-15, normal, cyclic subgroup $\tilde{R} \cong R_3 \times R_5 <R$, and (recognizing $R_3$ as a subgroup) $R_3 \; \mathrm{char}\; \tilde{R}$ (D&F, p.143).
Again since $\tilde{R} \lhd R$, we have (recognizing $R_3$ as a subgroup) $R_3 \lhd R$ by the same argument, also $R_3\; \mathrm {char}\; R$ (D&F, p.142).
And yet again, by $R \lhd G$, we get $R_3 \lhd G$.
Recognizing $R_3$ to be the only Sylow-3 subgroup, we reached contradiction as desired.
For part (c), as you said, first, we have to find possible image of $Z_4$ into $Z_4 \times Z_2$.
Using tuple to denote the generator of image of $Z_4$, and they are presented as generators within $Z_4 \times Z_2$.
We find they are:
$(0,0)$ for trivial image $0$, giving a direct product;
$(0,1), (2,1), (2,0)$ for image $Z_2$;
$(1,0), (1,1)$ for image $Z_4$.
Note some image are isomorphic, giving isomorphic semidirect product, for example, $(3,1)$ is not counted.
Second, similarly, for possible images of $Z_2 \times Z_2$ within $Z_4 \times Z_2$, we get:
$(0,0)$ for trivial image $0 \times 0$, giving a direct product;
$(0,1), (2,0), (2,1)$ for image $Z_2 \times 0$ (or isomorphic $0 \times Z_2$);
$(2,0)$ together with $(0,1)$ for image $Z_2 \times Z_2$.
