## Tag Archives: quaternion group

### The automorphism group of the quaterion group is isomorphic to Sym(4)

Prove that $\mathsf{Aut}(Q_8) \cong S_4$.

We already know that $|\mathsf{Aut}(Q_8)| = 24$ by counting the possible images of, say, $i$ and $j$. By a previous theorem, it suffices to show that no automorphism of $Q_8$ has order 6.

Let $\varphi$ be an automorphism of $Q_8$. Now $Q_8$ has 6 elements of order 4, one element of order 2, and one identity; $\varphi$ must fix the identity and the element of order 2, so that we can consider $\varphi$ as an element of $S_A$, where $A = \{i,j,k,-i,-j,-k\}$. Suppose now that some automorphism $\varphi$ has order 6. Now $\varphi(i) = a$, where $a \notin \{i,-i\}$, and $\varphi(a) = b$, where $b \notin \{i,-i,a,-a\}$. Clearly $\varphi(-i) = -a$ and $\varphi(-a) = -b$. If $\varphi(b) = i$, then $\varphi = (i\ a\ b)(-i\ -a\ -b)$ and $\varphi$ has order 3; thus $\varphi(b) = -i$, and we have $\varphi = (i\ a\ b\ -i\ -a\ -b)$. Now $\varphi^3 = (i\ -i)(a\ -a)(b\ -b)$. Note that $ia \in \{b,-b\}$. If $ia = b$, then $ia = (-i)(-a)$ $= \varphi^3(i)\varphi^3(a)$ $= \varphi^3(ia)$ $= \varphi^3(b) = -b$. Similarly, if $ia = -b$, then $ia = b$. Both cases yield a contradiction, so that no automorphism of order 6 exists.

In addition, we record this handy lemma.

Lemma: Let $G$ be a group and let $K \leq G$ be characteristic. Then $\mathsf{Aut}(G)$ acts on $G/K$ by $\varphi \cdot xK = \varphi(x)K$. Proof: First we need to show well-definedness: if $xK = yK$, we have $xy^{-1} \in K$. Then $\varphi(xy^{-1}) = \varphi(x)\varphi(y)^{-1} \in K$ since $K$ is characteristic in $G$, so that $\varphi(x)K = \varphi(y)K$. Moreover, $1 \cdot xK = 1(x)K = xK$ and $(\varphi \circ \psi) \cdot xK = (\varphi \circ \psi)(x)K = \varphi(\psi(x))K$ $= \varphi \cdot (\psi(x)K)$ $= \varphi \cdot (\psi \cdot xK)$. $\square$

### Exhibit a presentation for the quaternion group

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

It suffices to show that some generating set of $Q_8$ satisfies these relations and that any group with this presentation has at most 8 elements.

Evidently, letting $i = a$ and $j = b$, we have $i^2 = -1 = j^2$ and $(-i)ji = (-i)(-k) = -j$.

Now consider the presentation $\langle a, b \ |\ a^2 = b^2, a^{-1}ba = b^{-1} \rangle$. Note that $a^2b^{-1} = 1$, so that $a^2(a^{-1}ba)(a^{-1}ba) = 1$, so that $ab^2a = 1$. From this we deduce that $a^4 = 1$ and thus $b^4 = 1$.

• There is one reduced word of length 0: 1.
• There are two reduced words of length 1: $a$ and $b$.
• There are at most four reduced words of length 2: $a^2$, $ab$, $ba$, and $b^2 = a^2$. We see that at most three of these are reduced.
• There are at most six reduced words of length 3: $a^3$, $a^2b$, $aba = b$, $ab^2 = a^3$, $ba^2 = a^2b$, and $ba^2 = a^2b$. We see that at most two of these are reduced.
• There are at most four reduced words of length 4: $a^4 = 1$, $a^3b = ba$, $a^2ba = ab$, and $a^2b^2 = 1$. None of these are reduced.

Thus in fact $Q_8 = \langle a, b \ |\ a^2 = b^2, a^{-1}ba = b^{-1} \rangle$.

### Compute presentations for a given central product of groups

Give presentations for the groups $Z_4 \ast_\varphi D_8$ and $Z_4 \ast_\psi Q_8$ constructed in a previous example.

Note that $Z_4 \times D_8$ is generated by $(x,1)$, $(1,r)$, and $(1,s)$. Hence $Z_4 \ast_\varphi D_8$ is generated by the images of these elements under the natural projection; moreover, the relations satisfied by these also hold in the quotient. We have an additional relation: note that $(x,1)^2(1,r)^2 = (x^2,r^2) \in Z$. Thus $Z_4 \ast_\varphi D_8 = \langle a,b,c \ |\ a^4 = b^4 = c^2 = 1, a^2 = b^2,$ $ab = ba, ac = ca, bc = cb^3 \rangle$.

Similarly, $Z_4 \ast_\psi Q_8$ is generated by the natural images of $(x,1)$, $(1,i)$, and $(1,j)$, and we have an additional relation because $(x,1)^2(1,i)^2 = (x^2,-1) \in Z$. Thus $Z_4 \ast_\psi Q_8 = \langle a,b,c \ |\ a^4 = b^4 = c^4 = 1,$ $a^2 = b^2 = c^2, ab = ba, ac = ca, bc = cb^3 \rangle$

### Basic properties of the central product of groups

Let $A$ and $B$ be groups. Suppose $Z_1 \leq Z(A)$ and $Z_2 \leq Z(B)$ are subgroups and that there exists an isomorphism $\varphi : Z_1 \rightarrow Z_2$. Define $Z \leq A \times B$ by $Z = \{ (x, \varphi(x)^{-1}) \ |\ x \in Z_1 \}$. Note that $Z$ is normal in $A \times B$ since $Z \leq Z(A \times B)$, using a previous theorem. We denote by $A \ast_\varphi B$ the quotient $(A \times B)/ Z$. (In particular, $A \ast_\varphi B$ depends on $\varphi$.) Think of $A \ast_\varphi B$ as the direct product $A \times B$ “collapsed” by identifying each element $x \in Z_1$ with its $\varphi$-image in $Z_2$.

• Prove that the images of $A$ and $B$ in $A \ast_\varphi B$ are isomorphic to $A$ and $B$, respectively, and that these images intersect in a central subgroup isomorphic to $Z_1$. Find $|A \ast_\varphi B|$.
• Let $Z_4 = \langle x \rangle$. Let $D_8 = \langle r,s \rangle$ and $Q_8 = \langle i,j \rangle$ as usual. Let $Z_4 \ast_\varphi D_8$ be the central product of $Z_4$ and $D_8$ which identifies $x_2$ and $r^2$. (I.e. $Z_1 = \langle x^2 \rangle$, $Z_2 = \langle r^2 \rangle$, and $\varphi(x^2) = r^2$.) Let $Z_4 \ast_\psi Q_8$ be the central product of $Z_4$ and $Q_8$ which identifies $x^2$ and $-1$. (I.e. $Z_1 = \langle x^2$, $Z_2 = \langle -1 \rangle$, and $\psi(x^2) = -1$.) Prove that $Z_4 \ast_\varphi D_8 \cong Z_4 \ast_\psi Q_8$.

1. Let $\pi : A \times B \rightarrow (A \times B)/Z$ denote the canonical projection, and identify $A$ with $A \times 1$ and $B$ with $1 \times B$ in $A \times B$.

Let $(a_1,1), (a_2,1) \in A \times 1$, and suppose $\pi((a_1,1)) = \pi((a_2,1))$. Then $(a_2a_1^{-1},1) \in Z$, so that $a_2a_1^{-1} = \pi(1) = 1$. Thus $a_1 = a_2$, and hence $\pi|_A$ is injective. Similarly, $\pi|_B$ is injective.

Note that the restriction $\pi|_{Z_1}$ is also injective. Suppose $(x,y)Z \in \pi[A] \cap \pi[B]$; then for some $a \in A$ and $b \in B$ we have $(x,y)Z = (a,1)Z = (1,b)Z$. Thus $(a,b^{-1}) \in Z$; by definition, then, $a \in Z_1$ and $b = \varphi(a)$. Note that for all $(z,w)Z \in A \ast_\varphi B$ we have $(z,w)Z(a,1)Z = (za,w)Z$ $= (az,w)Z$ $(a,1)Z(z,w)Z$, so that $\pi[A] \cap \pi[B]$ is in the center of $A \ast_\varphi B$. Moreover, $(x,y)Z \in \mathsf{im}\ \pi|_{Z_1}$. Conversely, if $(z,1) ]in Z_1 \times 1$, we have $(z,1)Z = (z,1)(z^{-1}, \varphi(z))Z$ $= (1,\varphi(z))Z$, so that $\mathsf{im}\ \pi|_{Z_1} \subseteq \pi[A] \cap \pi[B]$. Then $\mathsf{im}\ \pi|_{Z_1} = \pi[A] \cap \pi[B]$, and by the First Isomorphism Theorem, $Z_1 \cong \pi[A] \cap \pi[B]$. In addition, $\pi[A] \cap \pi[B]$ is a central subgroup of $A \ast_\varphi B$.

Finally, by Lagrange write $|A| = n|Z_1|$ and $|B| = m|Z_2|$. Note that $|Z_1| = |Z_2|$. Now $|A \ast_\varphi B| = |A \times B|/|Z|$ $= nm|Z_1|^2/|Z_1|$ $= nm|Z_1|$. We may also write this equation in the form $|A \ast_\varphi B| = |A| \cdot [B : Z_2] = |B| \cdot [A : Z_1]$.

2. Define $\overline{\alpha} : \{(x,1), (1,i), (1,j)\} \rightarrow Z_4 \times D_8$ as follows: $\overline{\alpha}((x,1)) = (x,1)$, $\overline{\alpha}((1,i)) = (1,r)$, and $\overline{\alpha}((1,j)) = (x,s)$. Because $\{(x,1),(1,i),(1,j)\}$ generates $Z_4 \times Q_8$ and these images satisfy the relations $\overline{\alpha}((x,1))^4 = \overline{\alpha}((1,i))^4 = \overline{\alpha}((1,j))^4 = 1$, $\overline{\alpha}((x,1))\overline{\alpha}((1,i)) = \overline{\alpha}((1,i))\overline{\alpha}((x,1))$, $\overline{\alpha}((x,1))\overline{\alpha}((1,j)) = \overline{\alpha}((1,j))\overline{\alpha}((x,1))$, and $\overline{\alpha}((1,i))\overline{\alpha}((1,j)) = \overline{\alpha}((1,j))\overline{\alpha}((1,i))^3$, $\overline{\alpha}$ extends to a homomorphism $\alpha : Z_4 \times Q_8 \rightarrow Z_4 \times D_8$.

If we let $\pi$ denote the natural projection, $\pi \circ \alpha$ is a mapping $Z_4 \times Q_8 \rightarrow Z_4 \ast_\varphi D_8$. We claim that $Z \leq Z_4 \times Q_8$ is contained in the kernel of this mapping; it suffices to show this for the (unique) nonidentity element of $Z$, namely $(x^2,-1)$. In fact, $(\pi \circ \alpha)(x^2,-1)$ = $\pi(\alpha(x^2, i^2))$ $= \pi(\alpha((x,1)^2(1,i)^2)$ $= \pi(x^2,r^2) = 1$. Thus $Z \leq \mathsf{ker}(\pi \circ \alpha)$. By the remarks on page 100 in the text, there exists a unique group homomorphism $\theta : Z_4 \ast_\psi Q_8 \rightarrow Z_4 \ast_\varphi D_8$ such that $\theta \circ \pi = \pi \circ \alpha$, and in particular, $\theta(tZ) = (\pi \circ \alpha)(t)$.

We now show that $\mathsf{ker}\ \theta = 1$. Suppose $\theta((t,u)Z) = 1$. Then $\pi(\alpha(t,u)) = 1$; note that $(t,u) = (x^a,i^bj^c)$ for some integers $a,b,c$, and that $\alpha(t,u) = (x^{a+c},r^bs^c)$. Then $(x^{a+c},r^bs^c) \in \{ (1,1), (x^2,r^2) \}$.

If $(x^{a+c},r^bs^c) = (1,1)$, then $c \equiv 0$ mod 2 and $b \equiv 0$ mod 4. Now if $c \equiv 0$ mod 4, then $a \equiv 0$ mod 4 and $(t,u) = (1,1) \in Z$. If $c \not\equiv 0$ mod 4, then $a \equiv 0$ mod 2 and $(t,u) = (x^2,-1) \in Z$.

If $(x^{a+c},r^bs^c) = (x^2,r^2)$, then $c \equiv 0$ mod 2 and $b \equiv 2$ mod 4. If $c \equiv 0$ mod 4, then $a \equiv 2$ mod 4 and $(t,u) = (x^2,-1) \in Z$. If $c \not\equiv 0$ mod 4, then $a \equiv 0$ mod 4 and $(t,u) = (1,1) \in Z$.

Thus $\mathsf{ker}\ \theta = 1$, hence $\theta$ is injective.

Now by part (a), we see that both $Z_4 \ast_\varphi D_8$ and $Z_4 \ast_\psi Q_8$ have order 16. Thus $\theta$ is an isomorphism.

### In the direct product of the quaternion group and an elementary abelian 2-group, all subgroups are normal

Prove that all subgroups of $Q_8 \times E_{2^k}$ are normal, where $E_{2^k} = Z_2 \times \cdots \times Z_2$ is the elementary abelian group of order $2^k$.

Let $H \leq Q_8 \times E_{2^k}$ be a subgroup, and let $\alpha = (a,x) \in H$ and $\beta = (b,y) \in Q_8 \times E_{2^k}$. Since $E_{2^k}$ is abelian, we have $\beta\alpha\beta^{-1} = (bab^{-1}, x)$.

We saw previously that the conjugacy classes of $Q_8$ are $\{1\}$, $\{-1\}$, $\{i,-i\}$, $\{j,-j\}$, and $\{k,-k\}$. In particular, we have $bab^{-1} \in \{a, a^3\}$.

If $bab^{-1} = a$, then $\beta\alpha\beta^{-1} = (a,x) \in H$. If $bab^{-1} = a^3$, then (since every element of $E_{2^k}$ has order 2) $\beta\alpha\beta^{-1} = (a^3,x)$ $= (a^3, x^3)$ $= (a,x)^3 \in H$. Thus $H$ is normal.

### Q(8) has at most 24 distinct automorphisms

Show that $|\mathsf{Aut}(Q_8)| \leq 24$.

Recall that $Q_8 = \langle i,j \rangle$, and let $\varphi$ be an automorphism of $Q_8$.

We have $|i| = |j| = 4$. Then $|\varphi(i)| = |\varphi(j)| = 4$. There are 6 elements of order 4 in $Q_8$, so we have 6 choices for $\varphi(i)$. Note that if we choose $\varphi(j) \in \langle \varphi(i) \rangle$, then $\varphi$ is not surjective, hence not an isomorphism. Two of the six elements of order 4 are in $\langle \varphi(i) \rangle$, so we have 4 choices for $\varphi(j)$. Now $\varphi$ is determined by its image at $i$ and $j$, so that $|\mathsf{Aut}(Q_8)| \leq 24$.

### Compute the conjugacy classes of Dih(8), Alt(4), and the quaternion group

Find all conjugacy classes and their orders in the following groups: $D_8$, $Q_8$, and $A_4$.

1. In this previous exercise, we computed that $Z(D_8) = \{ 1, r^2 \}$. Thus each of $1$ and $r^2$ is in its own conjugacy class.

Now $\langle r \rangle \leq C_{D_8}(r) \leq D_8$, and the centralizer of $r$ is not all of $D_8$ since $rs \neq sr$. Now $4$ divides $|C_{D_8}(r)|$, which properly divides 8 by Lagrange’s Theorem. So $\langle r \rangle = C_{D_8}(r)$. Thus $[D_8 : \mathsf{stab}(r)] = 2$, and so the conjugacy class containing $r$ has order 2. Moreover, $(sr)r(sr)^{-1} = sr^2r^{-1}s^{-1} = srs = r^3$. Hence latex \{r,r^3\}\$ is a conjugacy class.

Now $\langle s, r^2 \rangle \leq C_{D_8}(s) \leq D_8$. The second inclusion is proper since $rs \neq sr$, and $|\langle s, r^2\rangle| \geq 4$. Thus $|C_{D_8}(s)| = 4$, and the conjugacy class containing $s$ has order 2. Moreover, $(sr)s(sr)^{-1} = srsr^3s = srs^2r = sr^2$. Hence $\{ s, sr^2 \}$ is a conjugacy class.

There are only two remaining elements, $sr$ and $sr^3$. Since $(r)sr(r)^{-1} = rs = sr^3$, $\{ sr, sr^3 \}$ is a conjugacy class.

Thus the conjugacy classes of $D_8$ are $\{1\}$, $\{r^2\}$, $\{r,r^3\}$, $\{sr,sr^3\}$, and $\{s,sr^2\}$, of order 1,1,2,2, and 2, respectively.

2. In this previous exercise, we found that $Z(Q_8) = \langle -1 \rangle$. We also know the subgroup lattice of $Q_8$; from this lattice and because $\langle x \rangle \leq C_{Q_8}(x)$ for all $x$, since every noncentral element generates a $\leq$-maximal subgroup, if $x \notin Z(Q_8)$ then $C_{Q_8}(x) = \langle x \rangle$. Thus each noncentral element is in a conjugacy class of order 2. Since $ji(-j) = -i$, $kj(-k) = -j$, and $ik(-i) = -k$, the conjugacy classes of $Q_8$ are $\{1\}$, $\{-1\}$, $\{i, -i\}$, $\{j, -j\}$, and $\{k, -k\}$, and have order 1,1,2,2, and 2, respectively.
3. $1$ is in its own conjugacy class in $A_4$.

In this previous exercise, we computed the subgroup lattice of $A_4$. If $\sigma = (a\ b\ c)$ is a 3-cycle in $A_4$, then $\langle \sigma \rangle$ is a maximal subgroup. Then $C_{A_4}(\sigma)$ is either $\langle \sigma \rangle$ or $A_4$; however, the centralizer of $\sigma$ is not all of $A_4$ because in particular $(a\ b)(c\ d) \in A_4$ but $(a\ b)(c\ d)(a\ b\ c) = (b\ d\ c)$ and $(a\ b\ c)(a\ b)(c\ d) = (a\ c\ d)$. Thus for any 3-cycle $\sigma$, $C_{A_4}(\sigma) = \langle \sigma \rangle$. Thus the conjugacy class containing any 3-cycle contains $[A_4 : \langle \sigma \rangle] = 4$ elements. Moreover, note that $(a\ b)(c\ d)(a\ b\ c)(a\ b)(c\ d) = (a\ d\ b)$, $(a\ c)(b\ d)(a\ b\ c)(a\ c)(b\ d) = (a\ c\ d)$, and $(a\ d)(b\ c)(a\ b\ c)(a\ d)(b\ c) = (b\ d\ c)$. Thus $\{(a\ b\ c), (a\ d\ b), (a\ c\ d), (b\ d\ c) \}$ is a conjugacy class. For different identifications of $a,b,c,d$ we have the conjugacy classes $\{(1\ 2\ 3), (1\ 4\ 2), (1\ 3\ 4), (2\ 4\ 3) \}$ and $\{ (1\ 3\ 2), (1\ 4\ 3), (1\ 2\ 4), (2\ 3\ 4) \}$.

Now let $\tau$ be any product of disjoint 2-cycles in $A_4$. From the subgroup lattice, the centralizer of $\tau$ is either $\langle \tau \rangle$, $A_4$, or the unique subgroup $K$ of order 4. Now the centralizer is not all of $A_4$ as before. Moreover, we saw in this previous exercise that the unique subgroup of order 4 is isomorphic to the Klein 4-group, hence is abelian. Thus $C_{A_4}(\tau) = K$. Thus the conjugacy class containing $\tau$ has $[A_4 : K] = 3$ elements. There are only three remaining elements; thus $\{ (1\ 2)(3\ 4), (1\ 3)(2\ 4), (1\ 4)(2\ 3) \}$ is a conjugacy class.

Thus the conjugacy classes of $A_4$ are $\{1\}$, $\{(1\ 2\ 3), (1\ 4\ 2), (1\ 3\ 4), (2\ 4\ 3)\}$, $\{(1\ 3\ 2), (1\ 4\ 3), (1\ 2\ 4), (2\ 3\ 4) \}$, and $\{(1\ 2)(3\ 4), (1\ 3)(2\ 4), (1\ 4)(2\ 3) \}$ of order 1, 4, 4, and 3, respectively.

### The quaternion group is not a subgroup of Sym(n) for any n less than 8

Let $Q_8$ be the quaternion group of order 8.

1. Prove that $Q_8$ is isomorphic to a subgroup of $S_8$.
2. Prove that $Q_8$ is not isomorphic to a subgroup of $S_n$ for any $n \leq 7$.

$Q_8$ is a subgroup of $S_8$ via the left regular representation.

Now suppose that $Q_8$ acts on a set $A$ of order at most 7. For any element $a \in A$, we have $[Q_8 : \mathsf{stab}(a)] = |Q_8 \cdot a| \leq 7$. By Lagrange’s Theorem, then, $|\mathsf{stab}(a)| \in \{2,4,8\}$. That is, $\mathsf{stab}(a)$ is nontrivial for all $a$.

We know that $\langle -1 \rangle$ is the unique $\leq$-minimal nontrivial subgroup of $Q_8$, so that $\langle -1 \rangle \leq \mathsf{stab}(a)$ for all $a \in A$. Thus if $K$ is the kernel of this action, $\langle -1 \rangle \leq K$ so that the action is not faithful.

Now if $Q_8$ is isomorphic to a subgroup of $S_7$, then the injection $Q_8 \rightarrow S_7$ induces a faithful action of $Q_8$ on the set $\{1,2,\ldots,7\}$. However, we showed that no such action can be faithful. Thus $Q_8$ is not isomorphic to a subgroup of $S_7$, and since $S_k \leq S_7$ for $1 \leq k \leq 7$, $Q_8$ is not isomorphic to a subgroup of $S_k$ for all $1 \leq k \leq 7$.

### Exhibit generators of an isomorphic copy of Q(8) in Sym(8) via the left regular representation

Use the left regular representation of $Q_8$ to produce two elements of $S_8$ which generate a subgroup isomorphic to $Q_8$.

Recall that $Q_8 = \langle i,j \rangle$.

Now $i(1) = i$, $i(-1) = -i$, $i(i) = -1$, $i(-i) = 1$, $i(j) = k$, $i(-j) = -k$, $i(k) = -j$, and $i(-k) = j$. Thus $i \mapsto (1\ i\ -1\ -i)(j\ k\ -j\ -k)$.

Similarly, $j(1) = j$, $j(-1) = -j$, $j(i) = -k$, $j(-i) = k$, $j(j) = -1$, $j(-j) = -1$, $j(k) = i$, and $j(-k) = -i$. Thus $j \mapsto (1\ j\ -1\ -j)(i\ -k\ -i\ k)$.

With the labeling

 $1 \mapsto 1$ $-1 \mapsto 2$ $i \mapsto 3$ $-i \mapsto 4$ $j \mapsto 5$ $-j \mapsto 6$ $k \mapsto 7$ $-k \mapsto 8$

we have $Q_8 \cong \langle (1\ 3\ 2\ 4)(5\ 7\ 6\ 8), (1\ 5\ 2\ 6)(3\ 8\ 4\ 7) \rangle$.

### Sym(4) has no subgroup isomorphic to the quaternion group

Prove that $S_4$ has no subgroup isomorphic to $Q_8$.

Suppose a subgroup $H \leq S_4$ exists such that $H \cong Q_8$.

We saw in a previous exercise that $Q_8$ contains 6 elements of order 4. Now $S_4$ contains exactly 6 elements of order 4: the 4-cycles. thus $H$ contains all 4-cycles in $S_4$. But then $H$ also contains all products of two 2-cycles, so $|H| > 8$, a contradiction. Thus no such $H$ exists.