The Quaternion Group, Q₈

Theorem.  Let a and b be in a group G such that |a|=|b|=4, a²=b², ba=a³b=a^-1b.   Then

1. If |c|=4, then for any integer n, cⁿ = c^d where d=n mod 4 and d is in {0,1,2,3}.
2. For any integers n and j, aⁿb^j = aⁿ⁺²b^j-2 for any integers n and j (because a²=b² ).
3. For any integers j and k, (a^jb^k)^-1=b^3ka^3j.
4. For any integer m, b^ka^m=a^t b^k where t=3^km.
5. For any integers j, k, m and n, (a^jb^k)(a^mbⁿ)=a^sb^t where s=(j+3^km)mod 4 in [0,3] and t=(k+n) mod 4 in [0,3].  If t is 2 or 3, s and t can be replaced by s'=(j+3^km+2) mod 4 in [0,3] and t by t'=t-2 in [0,1].
6. b³a=ab and ab is distinct from ba.  Thus <a,b> is not abelian.
7. <a,b> ={a^jb^k | j in [0,3] and k in [0,1] } and hence <a,b> has at most 8 elements.
8. <a,b> has exactly 8 elements.
9. In <a,b>, e has order 1, a² has order 2, and the remaining six elements have order 4.

Proof.   Let e denote the identity of G.  To prove (1), let n be any integer.  Then aⁿ = a^4k+d = a^4k a^d =(a⁴)^k a^d = e^k a^d  = e a^d = a^d.  To prove (2), note that aⁿb^j = aⁿ⁺²b^j-2 for any integers n and j, because a²=b²:  aⁿb^j =
aⁿ b^2+(j-2) = aⁿ (b² b^j-2) = (aⁿ b²) (b^j-2) = (aⁿ a²) b^j-2 =aⁿ⁺² b^j-2.  To prove (3), note that |a|=|b|=4.  If e is the identity of G, then e=a⁴=b⁴ and hence a(a³)=b(b³)=e.  Thus a^-1=a³ and b^-1=b³.  Hence (a^jb^k)^-1=(b^k)^-1 (a^j)^-1 = b^-k a^-k = (b^-1)^k (a^-1)^k = (b³)^k (a³)^j = b^3k a^3j.

To prove (4), use several induction arguments. 

    First let k=0 and let m be arbitrary.  Then b⁰a^m = ea^m = a^m = a^m b⁰.  Note that m does indeed equal 3⁰m.  Next let k=1 and m=0.  Then b¹a⁰ = b¹e = b¹ = eb¹ = a^3*0 b¹.  Now suppose that b¹a^m=a^3mb¹ for some m>-1, and consider b^ka^m+1. 

    b¹a^m+1 = b¹(a^m a¹) = (b¹a^m)a = (a^3mb¹)a = a^3m(b¹a) = a^3m(ba)= a^3m(a³b) = a^3m+3b = a^3(m+1)b¹.

By the Principle of induction, (4) now holds for k=1 and non-negative integers m.  Continue to let k=1 and consider m negative.  Then m=-n for some positive integer m, and using (3) and what we just proved for non-negative m with k=1,

    b¹a^m=b¹ a^-n = b¹ (aⁿ)^-1 = b¹a³ⁿ = a^3*3n b¹.

Note that 3*3=1 mod 4 and thus 3*3n mod 4 = n mod 4.  Thus a^3*3n = aⁿ by (1).  Hence, using (3) again,

    b¹a^m = aⁿ b¹ = a^-m b¹ = (a^m)^-1 b¹ = a^3m b¹.

Thus (4) holds for k=1 and all integers m. 

   Next suppose that (4) holds for some k >0, with no restriction on m.  Consider a case of k+1 and an arbitrary m.  Then, applying the case of k (to get w=3^km) and the case of k=1,

    b^k+1a^m = (b¹ b^k)a^m = b¹(b^ka^m) =b¹(a^wb^k) =(b¹a^w)b^k = (a^3w b¹) b^k = a^3w (b¹ b^k) = a^3w b^k+1.

Note that 3w =3*(3^km)=3^k+1m, as required in (4).  Thus, by Principle of Induction (and the previous cases done for k=0 and k=1), (4) holds for all non-negative k and arbitrary integers m.

    Finally, consider cases of  (4) with k<0.  Then k=-x for some x>0.  Apply (3) above to obtain b^k = b^-x = (b^x)^-1 = b^3x.  Then

    b^ka^m = b^3x a^m = a^w b^3x = a^w b^k, with w=3^3xm.

Note that 3²=1 mod 4,  and thus 3^g=3^{[g mod 2]} mod 4 for any integer g (see the proof of (1), and modify for order 2 instead of order 4).  Also, 3 mod 2 = 1 mod 2 = -1 mod 2, and thus  3x mod 2 = (3 mod2) (x mod 2) =(-1 mod 2) (x mod 2) = -x mod2.  Hence
            w mod 4 = (3^3xm) mod 4
                           = (3^3x mod 4) (m mod 4)
                           = (3^-x mod4) (m mod 4)
                           = (3^-xm) mod 4
                           = (3^km) mod 4. 
Thus a^w = a^s where s=3^km, by (1).  Thus, b^ka^m=a^sb^k with s=3^km as required by (4).  That completes the proof of (4).

    To prove (5), apply (4).  (a^jb^k)(a^mbⁿ) = [a^j (b^k a^m)] bⁿ = [a^j (a^w b^k)] bⁿ

with w=3^km by (4).  Hence, by the associative law and laws of exponents and (1) above,

    (a^jb^k)(a^mbⁿ) = a^(j+w) b^(k+n) = a^[(j+w)mod4]b^[(k+n)mod4] with w=3^km.

Let s=(j+w) mod 4 and t=(k+n) mod 4.  If t=2 or t=3, use (1) and (2) above and the fact that
[(x mod 4)+y)] mod 4 = (x +y) mod 4.

     (a^jb^k)(a^mbⁿ) = a^s+2 b^t-2 = a^{[(s+2)mod 4]}b^t-2 = a^{[(j+w+2)mod4]}b^t-2, with t-2 in {0,1} and w=3^km.

    To prove (6), use a²=b², (5), the associative law and the laws of exponents.  b³a = (bb²)a = (ba²)a = b(a³) = (a⁰b¹)(a³b⁰) = a^sb^t, where s=(0+3*3)mod 4 = 1 mod 4 and t=(1+0)mod 4 = 1 mod 4.  Thus b³a =a¹ b¹ = ab.  Suppose that ba=ab.  Then b³a=ba (since b³a=ab).  Cancelling a and then b on the right, one finds that b²=e and hence |b|<4.  Since |b|=4, ba and ab are different.

    To prove (7), use (3) and (5).  Let S be the subset of G consisting of all elements of the form a^jb^k, for j in [0,3] and k in [0,2].  Clearly a and b are in S because a=a¹b⁰ and b=a⁰b¹ .  By (3) and (5) above, S is a subgroup of G.  Because <a,b> is the smallest such group, <a,b> is a subset of S and hence <a,b>=S.  Thus G has at most 8 elements, because there are 4 choices for j and two choices for k in describing the set S.

    To prove (8), note that both <a> and <b> are subsets of <a,b>.  Since |a| =4 and |b|=4, Theorem 3.4 on page 35 says that <a> has the four distinct elements a, a², a³, and e while <b> has the four distinct elements b, b², b³ and e.   If b is not in <a>, then {e, a, a2, a3, b} is a subset of <a,b> with 5 distinct elements.  Thus 4<|<a,b>|<9.  By Lagrange's Theorem, Corollary 4.6 on page 39, |<a>| divides |<a,b>| because <a> is a subgroup of <a,b>.  Since |<a>|=|a|=4, <a,b> must be a multiple of 4.  The only multiple of 4 in (4, 9) is 8.  Thus |<a,b>| = 8, provided of course that we prove that b is not in <a>.  We know that b is distinct from a (by hypothesis), and that b is distinct from e and a²=b² (because e, b, b² and b³ are distinct).  Suppose b=a³.  Then b²=a³b=ba, which implies b=a.  Thus, b is distinct from a³ as well, and hence b is not in <a>.

    To prove (9), note first that |e|=1 always (for the identity of any group).  Also, since |a|=4, a² is distinct from e but a²a²=a⁴=e; thus |a²|=2.  The generators a and b have order 4 by hypothesis; this applies to a^-1=a³ and b^-1=b³=a²b as well, as a consequence of Theorem 3.6 on page 36.  In general, if c has order 4, so does c^-1.  Note that -1 and 4 are relatively prime; or, if you prefer, c^-1=c³ with 3 and 4 relatively prime.  By Theorem 3.6, c³ generates <c> and hence |c³|=|<c>|=4.  Consider next ab:  (ab)²=a^(1+3*1)mod4b^(1+1)mod4  = a⁰b²=a²; (ab)³=(ab)²(ab)=a²(ab)=(a² a)b=a³b; (ab)⁴=(ab)²(ab)²=a²a²=e.  Thus, |ab|=4 by Theorem 3.4 on page 35.  This applies as well to (ab)^-1=(ab)³=a³b.  Thus, a, a³=a^-1, b, a²b=b³=b^-1, ab and a³b=(ab)^-1 have order 4.

QED

Example.  Let A be the 2 by 2 matrix {{0,1},{-1,0}} (listed as rows) and B be the matrix {{0,i},{i,0}}, with complex numbers as entries.  Note that A and B are distinct and that, under the usual matrix multiplication, |A|=|B|=4, A²=B² and BA=A³B=A^-1B.

1. A²={{-1,0},{0,-1}}; A³={{0,-1},{1,0}}; A⁴={{1,0},{0,1}}=E, the identity for the 2 by 2 matrices under multiplication.  By Theorem 3.4, |A|=4.
2. B²={{-1,0},{0,-1}}=A²; B³={{0,-i},{-i,0}}; B⁴={{1,0},{0,1}}=E.  Thus |B|=4 and B²=A².
3. BA = {{-i, 0},{0,i}}; A³B= {{-i,0},{0,i}}.   Thus BA=A³B.

It is well-known that the set W is a group under matrix multiplication, where W consists of 2 by 2 matrices with non-zero determinants.  Thus A and B generate a subgroup <A,B> of W.  By the previous lemma, <A,B> has exactly 8 elements:  {E, A, A², A³, B, AB, A² B, A³ B}.  Here is the list in matrix form:

It is clear that <A,B> is isomorphic to any group <a, b> that meets the conditions of the previous theorem:  part (5) of that theorem applies to products (A^jB^k)(A^mBⁿ) and (a^jb^k)(a^mbⁿ) to produce homomorphic results if one simply maps A^jB^k to a^jb^k for all j in {0,1,2,3} and k in {0,1}.

Multiplication Table.  Below is a multiplication table for the group described above.

Proper Subgroups.  The multiplication tables of the four proper subgroups of <A,B> are color-coded above:

1. The table for {E, A²} is in green.
2. The table for {E, A, A², A³} is in yellow and green.
3. The table for {E, B, B²=A², B³=A²B} is in purple and green.
4. The table for {E, AB, A², A³B} is in orange and green.

Why are there only these 4 proper subgroups?  By Lagrange's Theorem, a proper subgroup can have 2 elements or 4.  The only element of order 2, with which to generate a subgroup of 2 elements, is A².  A subgroup H of order 4 must contain something outside of {E,A²}, say some C.  Since all elements except E and A² have order 4, H contains <C> and |<C>|=4.  Thus H=<C>.  There are six choices for C:  A and A³ give (2) above; B and B³=A²B give (3) above; AB and A³B give (4) above.

All Proper Subgroups Are Normal ({E} and <A,B> are always normal in <A,B>).  Any subgroup with 4 elements has index 2 in <A,B>.  Subgroups of index 2 are always normal (Exercise 1, page 45).  For any x in <A,B>, if we conjugate {E,A²} by x we get an isomorphic subgroup, x{E,A²}x^-1 (Exercise 6, page 45).  However, there is only one subgroup that consists of 2 elements:  {E, A²}.  Thus x{E,A²}x^-1 = {E, A²} for all x in <A,B> and {E,A²} is hence normal in <A,B>.

The Lattice Structure of Subgroups.  {E} is below {E,A²}, which is below all the rest.  The meet of any two of (2), (3) and (4) above is {E,A²}.  The join of any two of (2), (3) and (4) above is <A,B>.  Of course, <A,B> is above all the rest.

The Center of <A,B> is {E,A²}.  From the multiplication table above, note that B, AB, A²B and A³B do not commute with A.  Also, A³ does not commute with B.  However, A² commutes with everything.

The Ascending Central Series.  C₁(<A,B>) is the center of <A,B>, {E,A²}.  C₂(<A,B>) is the canonical pull-back to <A,B> of the center of <A,B>/{E,A²} (the center of a group is always a normal subgroup).  However, <A,B>/{E,A²} is a group of 4 elements and hence isomorphic to Z₄ or to (Z₂)x(Z₂) (see Exercise 5, page 40).  Hence this quotient group is abelian and equal to its center.  Thus C₂(<A,B>)=<A,B>.  Just for the record, <A,B>/{E,A²} is isomorphic to (Z₂)x(Z₂), because a²=b²=(a³)²=(a²b)²=(ab)²=(a³b)² for all six elements of <A,B> that are outside {E,A²}.

An Alternate Description of Q₈.  Let 1 be the identity, I=A, J=B and K=AB.  Let -1=A², -I=A³, -J=A²B, and -K=A³B.  Then

• -1*1=-1, -1*I=-I, -1*J=-J, -1*K=-K, (-1)*(-1)=1, (-1)*(-I)=I, (-1)*(-J)=J, (-1)*(-K)=K.  Thus multiplication by 1 or -1 acts like scalar multiplication by 1 or -1, respectively, on vectors.
• Since A² commutes with all of <A,B>, -1 freely factors from any product:  [(-1)x]y=(-1)(x*y), x*[(-1)y]=(-1)(xy), [(-1)x][(-1)y]=xy.  This is exactly how scalar multiplication works among cross products of vectors in R³.
• IJ=K; JK=I and KI=J.  Thus I, J, and K act like the cross-products of the standard basis vectors in R³.
• JI=-K, KJ=-I, and IK=-J.  This too copies the behavior of the cross-products of the standard basis vectors in R³.
• I²=J²=K²=-I.  This does not copy the behavior of cross products; for vectors in R³, the cross of v with itself is 0.

    Conversely, suppose a group H has two distinct generators a and b, each of order 4, such that a²=(ab)².  The condition (ab)(ab)=a² implies that b(ab)=a and hence (ba)b=a.  Thus, ba=ab^-1=ab³.  Thus, any such group must be Q₈.