Exercise 1.5.3 (Presentation of $Q_8$)

Question

Find a set of generators and relations for $Q_8$.

Richard Ganaye · Accepted Answer

\begin{proof} 
As a first solution, we can take the $64$ relations given by the Cayley table of Exercise 2!!!
$$Q_8 \simeq \langle e, e', i, i', j, j', k, k' \mid ea = e, ei = i, \ldots, k'k' =e' angle,$$
where $e$ represents $1$, $i'$ represents $-i$, and so on.

This shows that every finite group can be defined by generators and relations. Let us call this presentation the trivial presentation.

\bigskip
A more compact solution is given in Ex. 6.3.7 (p. 220):
$$Q_8 \simeq \langle a, b \mid a^2 = b^2,\ a^{-1} b a = b^{-1} angle.$$
First note that $Q_8 = \langle i, j angle$ satisfies the relations of this presentation for $i = a,\ j =b$, since $a^2 = i^2 = -1 = j^2 = b^2$, so that $a^2 = b^2$, and $jij = jk = i$, thus $i^{-1}j i = j^{-1}$, that is $a^{-1} b a = b^{-1}$.

Conversely, consider  the group $G = \langle a, b \mid a^2 = b^2,\ a^{-1} b a = b^{-1} angle$ with identity $e$ and generators $a,b$ such that $a^2 = b^2$ and $a^{-1} b a = b^{-1}$.

We prove first that $a^4 = e$. The relation $a^{-1} b a = b^{-1}$ gives $ bab =a$. Then
 $$a^2 = bab \cdot bab = b^2.$$
Multiplying by $b^{-1}$ on the left and right, we obtain $abba = e$, thus $ab = a^{-1}b^{-1}$ so $b^{2} = a^{-2}$. Therefore $a^4 =a^2 b^2 =a^2 a^{-2} = e$, which proves $a^4 = e$.

Since $ba = ab^{-1}$, and $a^4 = b^4 = e$, we can write every element of $G$ under the form $a^i b^j$ where $0 \leq i \leq 3,\ 0 \leq j \leq 3$. So every element of $G$ is in the list
\begin{align}
\begin{array}{cccc}
e, & a,& a^2, & a^3,\
b, & ab, & a^2 b, & a^3 b,\
b^2, & ab^2, &a^2 b^2, & a^2 b^3,\
b^3, & a b^3, & a^2 b^3, & a^3 b^3.
\end{array}
\end{align}
But $b^2 = a^2$, thus 
\begin{align}
\begin{array}{cccc}
b^2= a^2, & ab^2 = a^3, &a^2 b^2=e, & a^2 b^3 =a,\
b^3 =a^2 b, & a b^3=a^3b, & a^2 b^3 = b, & a^3 b^3 = ab.
\end{array}
\end{align}

Therefore
\begin{align}
G = \{e,a,a^2,a^3, b, ab, a^2 b, a^3 b\}.
\end{align}
(At this stage, we are not certain that these elements are distinct.)

Put 
\begin{align}
\begin{array}{llll}
\quad 1 = e,&\quad  i = a,& \quad j = b,&\quad  k =ab,\
\,-1 = a^2 = b^2,&\, -i = (-1) i = a^3, &\, -j = (-1)j = b^3,&\,  -k = (-1)k = a^3 b.
\end{array}
\end{align}
(Here $-1$ is only a writing of $a^2$, and not an algebraic relation.)

We verify all the relations defining $Q_8$ given in section 1.5.

First $1 \cdot x = x\cdot 1 = x$ for all $x \in G$, since $1 = e$ is the identity in $G$.

Since $a^4 = e$, $(-1)(-1) =a^4 = 1$, and by (2),  (3) and (4), for all $x \in G$, $(-1)\cdot x = a^2 x = -x.$
Note that $a^2$ commute with $a$, and since $a^2 = b^2$, $a^2$ commute with $b$, thus $a^2$ commute with all elements of $G = \langle a, b angle$. Therefore
$$\forall x \in G,\ (-1) x = x(-1) = x.$$

Moreover, we may verify with (4) and the relations $a^2 = b^2,\ a^{-1} b a = b^{-1} $ that
\begin{align*}
&i^2 = j^2 = k^2 = -1,\
&ij = k,\qquad ji = -k,\
&jk =i,\qquad kj = -i,\
&ki = j,\qquad ik = -j.
\end{align*}
For instance, $j k = bab = a = i,$ and so on.

This shows that the $64$ relations of the trivial presentation are consequences of $a^2 = b^2,\ a^{-1} b a = b^{-1}$. Therefore these two presentations are equivalent, and
$$Q_8 \simeq \langle a, b \mid a^2 = b^2,\ a^{-1} b a = b^{-1} angle.$$
\end{proof}
Note: for a more formal and more compact proof, see the solution of Ex. 6.3.7, that I propose here at the same time.

Exercise 1.5.3 (Presentation of $Q_8$)

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Exercise 1.5.3 (Presentation of $Q_8$)

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