Each extra-special group of order $2^{2n+1}$ is a central product of $D_8$s or of $D_8$s and a single $Q_8$. This is Exercise 5.3.7(i) of Robinson's, "A Course in the Theory of Groups (Second Edition)". According to this search, it is new to MSE.
This is a classification problem. This Wikipedia entry describes the result.
The Details:
Let $p$ be prime. A $p$-group is a group all of whose elements have order $p$.
Let $G$ be a finite $p$-group. Then $G$ is extra-special if $G'$ (the derived subgroup of $G$) and $Z(G)$ (the centre of $G$) coincide and have order $p$.
The quaternion group $Q_8$ is defined to be the group given by the presentation
$$\left\langle x,y\, \middle|\, x^{2^2}=1, y^2=x^2, y^{-1}xy=x^{-1}\right\rangle.$$
The dihedral group $D_8$ of order eight is the group given by the presentation
$$\left\langle r,s\,\middle|\,r^{2^2}, s^2, srs=r^{-1}\right\rangle.$$
Let $G$ be a group with normal subgroup $G_1,\dots, G_n$. Then $G$ is the central product $G_1\circ\dots\circ G_n$ of those normal subgroups if:

*

*$G=G_1G_2\dots G_n$,

*$[G_i, G_j]=1$ for $i\neq j$, and

*for all $i$, $$G_i\cap\prod_{j\neq i}G_j=Z(G).$$
Robinson claims that $Z(G_i)=Z(G)$.
The Question:
Paraphrased:

Consider an extra-special group $G$ of  order $2^{2n+1}$. Prove $G$ is a central product of the $D_8$s or a central product of $D_8$s and a single $Q_8$.

There is a hint (which I have paraphrased):

Prove that a central product of two $D_8$s is a central product of two $Q_8$s.

Thoughts:
I thought I'd rewrite the question in terms of group presentations, so I asked the following question:
Given $H=\langle X_H\mid R_H\rangle$ and $K=\langle X_K\mid R_K\rangle$, find a presentation for Robinson's $H\circ K$
It turns out that it is insufficient to know presentations of groups $H,K$ in order to find a presentation of $H\circ K$.

I don't know how to use the hint, let alone prove what it suggests I prove. My guess, though, is that the parity of $n$ determines whether there is a $Q_8$ term in the central product (since, if I'm right, we can replace an even number of $Q_8$s by the same number of $D_8$s in the central product). According to the Wikipedia entry (cited above), the number of $D_8$s in the central product does not depend on the parity of $n$.

If $G$ is an extra-special $2$-group of order $2^{2n+1}$, then $G'=Z(G)\cong \Bbb Z_2$.

Previous, relevant questions of mine include:

*

*Find the upper central series of $Q_{2^n}$.

*Showing ${\rm Aut}(D_{2^n})\cong{\rm Aut}(Q_{2^n})$ for $n\ge 4$.

*Showing ${\rm Aut}(Q_{2^n})\cong{\rm Hol}(\Bbb Z_{2^{n-1}})$ for $n>3$

Please help  :)
 A: Let $Z:=Z(G) = \langle z \rangle$, with $z$ of order $2$.
Choose two non-commuting generators $a,b$ of $G$. Then $[a,b]=z$, and $H:=\langle a,b \rangle$ is extraspecial of order $8$, and hence is isomorphic to $D_8$ or $Q_8$.
Now let $c$ be another generator of $G$, not in $H$. If $c \not\in C_G(a)$, then $[c,a]=z$, so $[cb,a]=1$. Similarly, if $c \not\in C_G(b)$, then $[ca,b]=1$. So we can replace $c$ by an element in the same coset $cH$, to get $c \in C_G(H)$.
It follows that $G = HC_G(H)$. In fact this is a central product, and you can check that $C_G(H)$ is extraspecial of order $2^{2n-1}$. So, by induction, $C_G(H)$ is a central produce of groups isomorphic to $D_8$ or $Q_8$, and hence the same applies to $G$.
Now it should be clear that the result to be proved follows from $D_8 \circ D_8 \cong Q_8 \circ Q_8$.
I can't really give advice on proving the hint. They are both groups of order $32$, so you just have to work out an isomorphism (and you can check it in GAP).
BTW, this result is proved in various textbooks on group theory.
