Exercise 6.3.7 (Presentation of $Q_8$)

Question

Prove that the following is a presentation for the quaternion group of order $8$:
$$Q_8 = \langle a, b \mid a^2 = b^2,\ a^{-1} b a = b^{-1}\rangle.$$

Richard Ganaye · Accepted Answer

\begin{proof} 
More precisely, we prove that
$$Q_8 \simeq \langle a, b \mid a^2 = b^2,\ a^{-1} b a = b^{-1}angle.$$
First note that $Q_8 = \langle i, j angle$, since $k = ij,\ -1 = i^2$ and $-i = (-1)i,\ -j = (-1)j$ and $-k = (-1)k$.

Consider the free group $F(S) = F(a,b)$ generated by $S = \{a,b\}$, an let $\varphi: \{a,b\} 	o Q_8$ defined by $\varphi(a) = i$ and $\varphi(b) = j$.

By the universal property of $F(S)$, there exists a homomorphism $\psi:F(a,b) 	o G$ such that $\psi(a) = i,\ \psi(b) = j$, i.e. such that the following diagram commutes:
\begin{center}
\begin{tikzpicture}
   
ode (A) at (3,2.5) {$F(a,b)$};
   
ode (B) at (3,1) {$Q_8$};
   
ode (C) at (1,2.5) {$S$};
   \draw[->] (A) edge node[right]{$\psi$} (B)  ;
   \draw[->] (C) edge node [above]{inc} (A);
   \draw[->] (C) edge node [left]{$\varphi\,$} (B);
\end{tikzpicture}
\end{center}

Let $G$ denote the group $\langle a, b \mid a^2 = b^2,\ a^{-1} b a = b^{-1}angle$. By definition, $G = F(a,b)/ N$, where $N$ is the intersection of all normal subgroups of $F(a,b)$ containing $R = \{a^2b^{-2}, a^{-1}bab\}$.

By the definition of $Q_8$, we know that
$$i^2 = j^2( =-1),\qquad 	ext{and}\qquad i^{-1} j i = (-i)(-k) = ik =-j = j^{-1},$$
thus $\psi(a^2 b^{-2}) = \psi(a)^2 \psi(b)^{-2} =i^2j^{-2} = 1$ and $\psi(a^{-1}bab) = \psi(a)^{-1} \psi(b) \psi(a) \psi(b) = i^{-1} j i j = 1$, so
$$R = \{a^2b^{-2}, a^{-1}bab\} \subset \ker(\psi).$$
Since $\ker(\psi)$ is a normal subgroup of $F(a,b)$ which contains $R$, then $N \subset \ker(\psi)$ by definition of $N$.

Let $\pi : F(a,b) 	o G = F(a,b)/ N$ be the canonical projection defined by $\pi(g) = gN$, so that $\ker(\pi) = N$. Then
$\ker(\pi) \subset \ker(\psi)$, therefore there exists a homomorphism $\xi : G 	o Q_8$ such that $\xi \circ \pi = \psi$, i.e. the following diagram commutes:

\begin{center}
\begin{tikzpicture}
   
ode (A) at (3,2.5) {$F(a,b)$};
   
ode (B) at (3,1) {$Q_8$};
   
ode (C) at (1,2.5) {$S$};
   
ode (D) at (5,2.5) {$G$};
   \draw[->] (A) edge node[right]{$\psi$} (B)  ;
   \draw[->] (C) edge node [above]{inc} (A);
   \draw[->] (C) edge node [left]{$\varphi\,$} (B);
   \draw[->] (A) edge node [above]{$\pi$}(D);
   \draw[->] (D) edge node[right]{$\, \xi$}(B);
\end{tikzpicture}
\end{center}  
Moreover, since $i,j$ are generators of $Q_8$ and
$$i = \psi(a) = \xi(\pi(a)),\qquad j = \psi(b) = \xi(\pi(b)),$$
then $\xi : G 	o Q_8$ is a surjective homomorphism. This shows that $G$ has at least $8$ elements.(*) Now we will prove that $G$ has at most $8$ elements.

\bigskip
As in Exercise 1.5.3, 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 \subset \{e,a,a^2,a^3, b, ab, a^2 b, a^3 b\}.
\end{align}
This shows that $G$ has at most $8$ elements. With the first part of the proof, this shows that 
$$|G| = 8 = |Q_8|.$$
Therefore the surjective homomorphism $\xi:G	o Q_8$ is an isomorphism, so
$$H_8 \simeq G = \langle a, b \mid a^2 = b^2,\ a^{-1} b a = b^{-1}angle.$$
\end{proof}
(*) This is a proof in a particular case of van Dyck's Theorem (see the note in Exercise 2.4.7).

Exercise 6.3.7 (Presentation of $Q_8$)

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Exercise 6.3.7 (Presentation of $Q_8$)

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