Mathematics 243 – Algebraic Structures, section 1 Selected Solutions – Problem Set 6 November 14, 2003

3.2/11. Let H be a subgroup of a group G, let a ∈ G be a fixed element, and let

K = {x ∈ G|x = aha^{−1}, some h ∈ H}

We must show K is a subgroup of G. First, e ∈ K, since e = aea^{−1}, and e ∈ H because H is a subgroup. Next, if x, y ∈ K, we must show that xy ∈ K. Saying x ∈ K means x = aha^{−1} for some h ∈ H. Similarly y ∈ K means y = aka^{−1} for some k ∈ H. Then

xy = (aha^{−1})(aka^{−1}) = ah(a^{−1}a)ka^{−1} = aheka^{−1} = ahka^{−1}

Since H is a subgroup and h, k ∈ H, we have hk ∈ H. Therefore xy ∈ K. Finally, we must show that if x ∈ H, then x^{−1} ∈ H too. But note by the “reverse order law”,

x^{−1} = (aha^{−1})^{−1} = (a^{−1})^{−1}h^{−1}a^{−1} = ah^{−1}a^{−1}

Since H is a subgroup of G and h ∈ H, h^{−1} ∈ H. This says x^{−1} ∈ K. We have verified all three of the conditions of Theorem 3.9 in the text, so K is a subgroup of G.

3.2/12. Let G be an abelian group (i.e. the group operation is commutative), and let H = {x ∈ G|h^{−1} = h}. In other words, H is the set of elements in G that are their own inverses. We want to show that H is a subgroup of G. First, since e^{−1} = e for the identity e in G, e ∈ H, so H is nonempty. Next, if h, k ∈ H, so h = h^{−1} and k = k^{−1}, then by the reverse order law and commutativity,

hk = h^{−1}k^{−1} = (kh)^{−1} = (hk)^{−1}.

Hence hk ∈ H (because it is its own inverse). Finally, since h^{−1} = h for all h ∈ H, every element in H has an inverse in H (itself!!). We have verified all three of the conditions of Theorem 3.9 in the text, so H is a subgroup of G.

3.12/17. Let G be a group, and let

Z(G) = {a ∈ G|ax = xa, all a ∈ G}

We must show Z(G) is a subgroup of G. First, e ∈ Z(G), since ex = xe = x. Next, if a, b ∈ Z(G), we must show that ab ∈ Z(G). Saying a ∈ Z(G) means ax = xa for all x ∈ G. Similarly b ∈ Z(G) means bx = xb for all x ∈ G. Then using associativity and these properties we have for any x ∈ G:

(ab)x = a(bx) = a(xb) = (ax)b = (xa)b = x(ab)

Hence ab also commutes with every x in G, so ab ∈ Z(G). Finally, we must show that if a ∈ Z(G), then a^{−1} ∈ Z(G) too. From the equation ax = xa, if we multiply on both sides on the left and right by a^{−1} we have a^{−1}(ax)a^{−1} = a^{−1}(xa)a^{−1}. Regrouping by associativity, and cancelling when we can this gives xa^{−1} = a^{−1}x for all x in G. Hence a^{−1} ∈ Z(G). We have verified all three of the conditions of Theorem 3.9 in the text, so Z(G) is a subgroup of G.

3.3/13a. First, we show by induction on n that

( \begin{matrix}cos θ & − sin θ \\ sin θ & cos θ \end{matrix} )^{n} = ( \begin{matrix}cos nθ & − sin nθ \\ sin nθ & cos nθ \end{matrix} )

for all n ≥ 1. The base case is completely obvious (nothing to prove). So assume that

( \begin{matrix}cos θ & − sin θ \\ sin θ & cos θ \end{matrix} )^{k} = ( \begin{matrix}cos kθ & − sin kθ \\ sin kθ & cos kθ \end{matrix} )

and consider

( \begin{matrix}cos θ & − sin θ \\ sin θ & cos θ \end{matrix} )^{k+1} = ( \begin{matrix}cos θ & − sin θ \\ sin θ & cos θ \end{matrix} )^{k} ( \begin{matrix}cos θ & − sin θ \\ sin θ & cos θ \end{matrix} )

By the induction hypothesis, this is

( \begin{matrix}cos kθ & − sin kθ \\ sin kθ & cos kθ \end{matrix} ) ( \begin{matrix}cos θ & − sin θ \\ sin θ & cos θ \end{matrix} ) =

( \begin{matrix}cos kθ cos θ − sin kθ sin θ & − cos kθ sin θ − sin kθ cos θ \\ cos kθ cos θ + sin kθ sin θ & cos kθ cos θ − sin kθ sin θ \end{matrix} )

Using the trig identities

cos(A + B) = cos(A) cos(B) − sin(A) sin(B), sin(A + B) = sin(A) cos(B) + sin(B) cos(A)

this last matrix equals

( \begin{matrix}cos(kθ + θ) & − sin(kθ + θ) \\ sin(kθ + θ) & cos(kθ + θ) \end{matrix} ) = ( \begin{matrix}cos(k + 1)θ & − sin(k + 1)θ \\ sin(k + 1)θ & cos(k + 1)θ \end{matrix} )

which is what we wanted to show. To be complete, to finish part a, we should also show

that the negative powers of ( \begin{matrix}cos θ & − sin θ \\ sin θ & cos θ \end{matrix} ) have the same form. To see this note by our

general formula for 2×2 inverse matrices, since det ( \begin{matrix}cos θ & − sin θ \\ sin θ & cos θ \end{matrix} ) = cos^{2} θ +sin^{2} θ = 1,

we have

( \begin{matrix}cos θ & − sin θ \\ sin θ & cos θ \end{matrix} )^{−1} = ( \begin{matrix}cos θ & sin θ \\ − sin θ & cos θ \end{matrix} ) = ( \begin{matrix}cos −θ & − sin −θ \\ sin −θ & cos −θ \end{matrix} )

because of the identities sin(−θ) = − sin θ and cos(−θ) = cos(θ). Hence the negative

powers of ( \begin{matrix}cos θ & − sin θ \\ sin θ & cos θ \end{matrix} ) will come from the matrices of the same form with negative

integer multiples of θ.

3.3/14 and 15. We can show both of these simultaneously by showing that the set G of all [a] in Z_{n} with gcd(a, n) = 1 is a group under multiplication mod n. We know that multiplication mod n is associative in all cases, so it is associative on the subset of classes which have multiplicative inverses. We are including exactly the elements in Z_{n} that do have multiplicative inverses, so every element of G has an inverse for multiplication and that inverse is also an element of G. The element [1] ∈ G is an identity for multiplication. So the only thing that must be proved is that G is closed under multiplication. If gcd(a, n) = gcd(b, n) = 1, then we know there exist equations am+nℓ = 1 = bp+nq for some integeres m, ℓ, p, q. If we multiply, we get

1 = (am + nℓ)(bp + nq) = (ab)pm + n(ℓp + qa + nℓq)

This implies that gcd(ab, n) = 1 also. Hence [ab] also has a multiplicative inverse in Z_{n}.