I want to show that the given $2$ groups are isomorphic to each other; that is, I want to show that: $$G=\langle x,y\mid x^4=1,x^2=y^2,y^{-1}xy=x^{-1}\rangle$$ is isomorphic to the quaternion group $Q$. In fact the first one is another representation of $Q$ so somehow it's obvious.

However, I want to strictly show it by creating an isomorphism from $G \to Q$ concretely but what is this mapping? Should this mapping send a generator of $G$ (i.e. $x,y$) to $i,j$ in $Q$? Should I use other methods (for example isomorphism theorems) to prove it?

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Yes, if you send $x$ to $i$ and $y$ to $j$, then you can work out the images of all of the other elements of $G$. You need to check that the images of the relations in $G$are satisfied by their images in $Q$. So you would need to check that $i^4=1$, $i^2=j^2$ and $j^{-1}ij=i^{-1}$. – Derek Holt Jun 9 '13 at 9:59

To complement what Derek Holt has written in his comment, once you have shown that $i, j$ satisfy the relations for $x$ and $y$, you now know that $Q$ is a homomorphic image of $G$.

Now you show that $G$ has at most $8$ elements, from which it follows that $G$ and $Q$ are isomorphic.

To show that $G$ has at most $8$ elements, note that every element can be written as $y^{b} x^{a}$, with $0 \le a < 4$, and $y \in \{ 0, 1 \}$. In fact if you have any product of $x$ and $y$, you can move the $y$ to the left using $$x y = y x^{-1}.$$ Then note that $x^4 = 1$ and $y^{2} = x^{2}$ ensure that you do not need powers of $x$ other than $1, x, x^{2}, x^{3}$, and powers of $y$ other than $1, y$.

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You're on the right track, yes. Set $H = \langle x, y \rangle$ (the free group on two generators $x$ and $y$), then $G = H/N$, where $N = (x^4, x^2y^{-2}, xyxy^{-1})$. Now set $f : H \to Q$ such that $f(x) = i$ and $f(y) = j$. This $f$ exists by the universal property of free groups. Since $i,j$ generate $Q$, $f$ is surjective.

The only thing left is computing $\mathrm{ker}(f)$; we actually want to prove that it's equal to $N$.

• $f(x^4) = i^4 = 1$, $f(x^2y^{-2}) = i^2(-j)^2 = (-1)^2 = 1$, and $f(xyxy^{-1}) = -ijij = -k^2 = 1$, so $N \subset \mathrm{ker}(f)$.
• I'll leave you as an exercise that $\mathrm{ker}(f) \subset N$ (it's the most tedious part).

So by general theorems, $N \subset \mathrm{ker}(f) \Rightarrow f$ induces a morphism $\bar f : H/N = G \to Q$, and $\mathrm{ker}(f) \subset N \Rightarrow \bar f$ is injective. Since $f$ was surjective, so is $\bar f$, and so $f$ is an isomorphism.

This method is a good general method to work with group presentations. Send the generators somewhere, and verify that the relations are still satisfied. The most difficult and tedious part is the last one: verifying that you don't introduce new relations (ie. $\mathrm{ker}(f) \subset N$).

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The two presentations are

$$G = \langle x, y \;|\; xxxx = 1, xx = yy, x = yx^{-1}y^{-1} \rangle$$ $$Q = \langle -1, i, j, k \;|\; (-1)^2 = 1, i^2 = j^2 = k^2 = ijk = -1 \rangle$$

We can try to transform one presentation into the other.

The first relation hints that $-1 = xx (= yy)$ and $i = x$, $j = y$.

Now notice that the $ijk = -1$ relation will allow us to recover $k$ as $k = i^{-1}j^{-1}(-1) = x^{-1}y^{-1}yy = x^{-1}y$.

So introducing the (redundant, hence unchanging) relations $z = xx$ and $w = x^{-1}y$ you can explicitly see how one turns into the other.

I think the other answers are better, but it is kind of fun and a useful skill to be able to work with concrete presentations.

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