What does it mean that in a factor/quotient group certain elements get "glued" together In these these notes on the generalized quaternion group it is written that:

[...] $Q_{2^n}$ is made by taking a cyclic group of order $2^{n-1}$ and a cyclic group of order $4$ and "glueing" them at their elements of order $2$ [...]

In connection with quotient groups I often read about this correspoding to some sort of "glueing". But what is the meaning of it, the interpretations of the quotient group that I heard of and can make sense are:
i) it is something like a low-resolution version of the group,
ii) it is built by collapsing certain elements, more formally for a homomorphism we compute with the inverse images/fibers instead of individual elements (for vector spaces this gives the nice interpretation as computing with affine spaces)
iii) all elements in the group that are factored out are identified with $1$ in the quotient group, so a factor group corresponds to demanding certain elements to vanish,
iv) contrary to subgroups, that are contained as a whole, we are looking for "subobjects" that are contained "section-wise" and could be divided out, i.e. in some sense a more "subtle way of containment".
So what is this "glueing together" interpretation, by the language I expect to have some geometric image!?
 A: If $f:X \to Y$ is an arbitrary mapping between non-empty sets, then elements $x_{1}$ and $x_{2}$ of $X$ are glued or identified by $f$ if $f(x_{1}) = f(x_{2})$. The image $f(X) \subset Y$ may be viewed as "$X$, possibly with some points identified".
Formally, define an equivalence relation on $X$ by $x_{1} \sim x_{2}$ if and only if $f(x_{1}) = f(x_{2})$. The equivalence classes are precisely the preimages of singletons, and $f$ induces a natural bijection from $X/{\sim}$ to $f(X)$.
In the setting you ask about, there's additional structure, and $f$ respects this structure. Particularly, the first homomorphism theorem for groups asserts that if $f:G \to H$ is a group homomorphism with kernel $K$, then $f$ naturally induces an isomorphism from $G/K$ to $f(G)$. The language and concept of gluing are quite naive and general, however.
A: In this context, glueing means taking a suitable quotient of a suitable semidirect product.
Explicitly, let $\langle a \rangle$ be of order $2^{n-1}$ and $\langle b \rangle$ be of order $4$. Take the semidirect product $H$ of $\langle a \rangle$ by $\langle b \rangle$ with the action
$$\tag{act}
a^{b} = b^{-1} a b = a^{-1}.
$$
In $H$, you can prove that the subgroup $N = \langle b^{-2} a^{2^{n-2}} \rangle$ has order $2$ and it is central. (Clearly $a^{2^{n-2}}$ and $b^{2}$ have order $2$. Also, it follows easily from (act) that $b^{-2} a b^{2} = b^{-1} a^{-1} b = a$, that is, $a b^{2} = b^{2} a$; and $b^{-1} a^{2^{n-2}} b = a^{-2^{n-2}} = a^{2^{n-2}}$, that is, $a^{2^{n-2}} b = b a^{2^{n-2}}$. So both $b^{2}$ and $a^{2^{n-2}}$ are central, and so is their product $b^{-2} a^{2^{n-2}}$.) 
Now you obtain $Q_{2^{n}}$ as the quotient $H/N$.
In particular in the quotient you have "glued together" the elements $b^{2}$ and $a^{2^{n-2}}$, in the sense that in $H/N$ you have
$$
b^{2} N = b^{2} (b^{-2} a^{2^{n-2}}) N = a^{2^{n-2}} N,
$$
where $(b^{-2} a^{2^{n-2}}) N = N$ as $b^{-2} a^{2^{n-2}} \in N$
A: Let us take a simple example. $G_1,G_2$ be cyclic groups of order $4$. Let $G_i=\langle x\rangle$ and $G_2=\langle y\rangle$. We can obtain many new groups from these two. One of them is direct product $G_1\times G_2$. 
In this direct product, $\langle x^2y^{-2}\rangle$ is a normal subgroup of order $2$. So if we factor $G_1\times G_2$ by $\langle x^2y^{-2}\rangle$, we get another group. So we obtained a new group from $G_1,G_2$ by process 
$$G_1, G_2 \rightsquigarrow G_1\times G_2 \rightsquigarrow \mbox{quotient  of } G_1\times G_2.$$
What is happening when we go from $G_1,G_2$ to the last quotient? 
The element $x^2y^{-2}$ (in middle step) is going to identity in quotient; which means $x^2$ is becoming equal to $y^2$ (in quotient), i.e. $\langle x^2\rangle$ becoming same as $\langle y^2\rangle$ (in quotient). 
This is exactly what we interpret as $\langle x^2\rangle$ and $\langle y^2\rangle$ are glued.

(Making this more precise is stated in other answers.)
