Groups of order $16$ with a cyclic quotient of order $4$

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Question: I am interested in

(a) listing the groups $G$ of order $16$ which have a cyclic quotient of order $4$;

(b) in each case knowing in how many essentially different ways this occurs (identifying cases where the kernels are in the same orbit under $\mathop{Aut}(G)$).


My working (edited after Derek Holt pointed out a silly mistake:

(A) Of the $14$ groups of order $16$ there are five abelian ones, of which four namely

$\mathbb{Z}_{16}$,

$\mathbb{Z}_8\oplus \mathbb{Z}_2$,

$\mathbb{Z}_4\oplus\mathbb{Z}_4$,

$\mathbb{Z}_4\oplus\mathbb{Z}_2\oplus\mathbb{Z}_2$

have such a quotient. It seems to me that in each case we get only one such quotient.

(B) A non-abelian group $G$ of order $16$ can only have such a quotient if its derived group $G'$ has order $2$ and $G/G'\simeq\mathbb{Z}_4\oplus\mathbb{Z}_2$.

Inspecting the list at https://groupprops.subwiki.org/wiki/Groups_of_order_16 it seems to me that the only possibilities are

SmallGroup(16,3),

the non-trivial semi-direct product of Z4 and Z4,

the group M16

I am not sure about the uniqueness of the quotients.

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I answered your question about uniqueness using computer calculations.

For the first two nonabelian example, $\mathtt{SmallGroup}(16,3)$ and $C_4 \rtimes C_4$, there are two normal subgroups with quotient $C_4$, but in both cases there is an automorphism of the group mapping one to the other, do they are (essentially) unique.

In the third example $M_{16}$, however, there are again two normal subgroups with quotient $C_4$, but this time all group automorphisms fix both subgroups, so the two subgroups are essentially distinct.

$M_{16}$ can be defined by the presentation $\langle x,y \mid x^8=y^2=1, y^{-1}xy=x^5 \rangle$, and the two normal subgroups are $\langle x^4,y \rangle$ and $\langle x^4,yx^2 \rangle$.In fact these two subgroups are not even isomorphic: the second one is cyclic but the first one not.