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.