I take it you want to show there *is* an isomorphism from the second group to the cyclic group of order $4$. So you want to find an element of order $4$ in the quotient group. That will have to come from an element of order $4$ in the big group. One such element is $(0,1)$. Now just check to see that that element indeed has order
$4$ in the quotient group.

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 $Z_4$.

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