Is a Normal Subgroup of GL(2,R). Apply the One-Step Subgroup Test to N = SL(2,R)

Home , Abelian group, Cyclic group, Dihedral group, Normal subgroup, Order (group theory), Subgroup

Handout #9 Math 545 Fall, 2010

1. Prove that SL(2,R) is a normal subgroup of GL(2,R). Apply the One-Step Subgroup Test to N = SL(2,R). N 6= ∅ since the identity matrix of G has its determinant equal to 1 and therefore belongs to N. Let A, B ∈ N. Show AB−1 ∈ N. It suﬃces to show that det(AB−) = 1. det(AB−1) = det A det(B−1) (since the function det is multiplicative) 1 1 = det A det B = 1. 1 (since A, B ∈ N) = 1. Now show that N is a normal subgroup of GL(2,R). Let A ∈ GL(2, R) and B ∈ N. Show ABA−1 ∈ N. It suﬃces to show that det(ABA−) = 1. −1 −1 1 det(ABA ) = det A det B det(A ) (since the function det is multiplicative) = det A det B det A = 1 det A.1. det A (since B ∈ N) = 1.

2. Viewing < 3 > and < 12 > as subgroups of Z, list all elements of the factor group < 3 > / < 12 >. Let m ∈< 3 >. Then m = 12q + r, where r = 0, 3, 6 or 9. =⇒ m+ < 12 >= r + 12q+ < 12 >= r+ < 12 > (since 12q ∈< 12 >). Thus < 3 > / < 12 >= {m+ < 12 > |m ∈< 3 >} = {r+ < 12 > |r = 0, 3, 6, 9} = {< 12 >, 3+ < 12 >, 6+ < 12 >, 9+ < 12 >}

3. What is the order of the element 14+ < 8 > of the factor group Z24/ < 8 >? Z24 = {0, 1, 2,..., 23} and < 8 >= {0, 8, 16}. Z24/ < 8 >= {< 8 >, 1+ < 8 >, 2+ < 8 > , 3+ < 8 >, 4+ < 8 >, 5+ < 8 >, 6+ < 8 >, 7+ < 8 >}. Note 14+ < 8 >= 6 + (8+ < 8 >). Thus 14+ < 8 >= 6+ < 8 >(since 8 ∈ < 8 >). Since 1(6+ < 8 >) = 6+ < 8 >6=< 8 >; 2(6+ < 8 >) = 12+ < 8 >= 4+ < 8 >6=< 8 >; 3(6+ < 8 >) = 18+ < 8 >= 2+ < 8 >6=< 8 > and 4(6+ < 8 >) = 24+ < 8 >=< 8 > (because 24 ≡24 0 ∈< 8 >). Hence |6+ < 8 > | = 4 and since 14+ < 8 >= 6+ < 8 >, we have |6+ < 8 > | = 4.

4. Suppose that G is a non-Abelian group of order p3 (where p is prime) and Z(G) 6= {e}. Prove that |Z(G)| = p. Since Z(G) is a subgroup of G and |G| = p3. =⇒ |Z(G)| = 1, p, p2, or p3 (Lagrange). Now prove: |Z(G)| = p2 and |Z(G)| = p3 are impossible. |Z(G)| = p3 =⇒ |Z(G)| = |G| =⇒ G is abelian, 2 |G| p3 a contradiction since G is non-Abelian and |Z(G)| = p =⇒ |G/Z(G)| = |Z(G)| = p2 = p =⇒ G/Z(G) is of prime order p and therefore cyclic =⇒ G is abelian, a contradiction since G is non-abelian. Hence |Z(G)| = p (since |Z(G)|= 6 1 from the given).

5. If |G| = pq, where p and q are not necessarily distinct primes, prove that |Z(G)| = 1 or pq. |G| = pq =⇒ |Z(G)| = 1, p, q, or pq. Prove: |Z(G)| = p and |Z(G)| = q are impossible. If |G| pq |Z(G)| = p then |G/Z(G)| = |Z(G)| = p = q. But then, since G/Z(G) is cyclic of prime order, |G/Z(G)| = 1. Thus q = 1, a contradiction since q > 1. Similarly, |Z(G)| = q =⇒ p = 1, a contradiction to the deﬁnition of p.

6. If M and N are a normal subgroups of G, then prove that M ∩ N is a normal subgroup of G. Since M and N are subgroups of G, M∩N is a subgroup of G (done in class). Need to prove M∩N is normal in G. Let g ∈ G and let n ∈ M ∩ N. Show gng−1 ∈ M ∩ N. Now n ∈ M ∩ N =⇒ n ∈ M and n ∈ N. Then gng−1 ∈ M and gng−1 ∈ N (since n ∈ M, n ∈ N and g ∈ G and M and N are normal in G). Thus gng−1 ∈ M ∩ N.

7. If N and M are normal subgroups of G, prove that NM is also a normal subgroup of G. First note that e = ee ∈ NM, since e ∈ N and e ∈ M. So NM 6= ∅. Let n1m1, n2m2 ∈ NM, where n1, n2 ∈ N and m1, m2 ∈ M. −1 −1 −1 −1 −1 Then (n1m1)(n2m2) = n1m1(m2) (n2) = n1n3m1(m2) (since N is normal in G, m1(m2) N = −1 −1 −1 −1 −1 Nm1(m2) =⇒ m1(m2) (n2) = n3m1(m2) , where n3 ∈ N) Thus (n1m1)(n2m2) =

1 −1 −1 n1n3m1(m2) ∈ NM, since n1n3 ∈ N and m1(m2) ∈ M. Hence, by One-Step Subgroup Test, NM is a subgroup of G. Now show NM is a normal subgroup of G. Let nm ∈ NM and g ∈ G. Prove g(nm)g−1 ∈ NM. g(nm)g−1 = g(nem)g−1 = g(ng−1gm)g−1 = (gng−1)(gmg−1) ∈ NM (since N and M are normal in G, gng−1 ∈ N and gmg−1 ∈ M).

8. List all elements of Z/ < 6 >. Z/ < 6 >= {a+ < 6 > |a ∈ Z}. Let a ∈ Z. Then, by the division algorithm, ∃q, r ∈ Z such that a = 6q + r, where r = 0, 1, 2, 3, 4 or 5. =⇒ a+ < 6 >= r + 6q+ < 6 >= r+ < 6 > (since 6q ∈< 6 >). Thus Z/ < 6 >= {a+ < 6 > |a ∈ Z} = {r+ < 6 > |r = 0, 1, 2, 3, 4, 5} = {< 6 > , 1+ < 6 >, 2+ < 6 >, 3+ < 6 >, 4+ < 6 >, 5+ < 6 >}

2 3 4 5 2 3 4 5 9. Consider the dihedral group of order 12, D6 = {e, a, a , a , a , a , b, ab, a b, a b, a b, a b}, and its 3 center, Z(D6) = {e, a }. Find the order of the element aZ(D6) in the factor group D6/Z(D6). 4 aZ(D6) = {a, a }. 2 2 2 5 (aZ(D6)) = a Z(D6) = {a , a }. 3 3 3 6 (aZ(D6)) = a Z(D6) = {a , a = e} = Z(D6). Thus |aZ(D6)| = 3. 10. Prove that a factor group of a cyclic group is cyclic. Let G be a cyclic group, say G =< a > and let H be a subgroup of G. Then H =< at >, where t ∈ Z (since every subgroup of a cyclic group is cyclic). Prove: G/H =< aH >. Let gH ∈ G/H. Then g ∈ G =⇒ g = as, for some s ∈ Z. Thus gH = asH = (aH)s. =⇒ gH ∈< aH >. 11. List all generators of the group Z/ < 9 >. Z/ < 9 >=< 1+ < 9 >> and |1+ < 9 > | = 9 . Now k+ < 9 >∈ Z/ < 9 > is a generator of Z/ < 9 > ⇐⇒ gcd(k,9) = 1. Since 1 ≤ k ≤ 9, the generators of Z/ < 9 > are 1+ < 9 >, 2+ < 9 >, 4+ < 9 >, 5+ < 9 >, 7+ < 9 >, and 8+ < 9 >.

12. Let G =< a > be a cyclic group of order 30 and H =< a5 >. List all generators of the group G/H. G/H =< a + H > and |a + H| = 5 (see problem 7 above). Now ak + H ∈ G/H is a generator of G/H ⇐⇒ gcd(k,5) = 1. Since 1 ≤ k ≤ 5, the generators of G/H are a + H, a2 + H, a3 + H, and a4 + H.