1. The odd get even

Let G be a subgroup of S_n. Show that if G contains an odd permutation, then |G| is even.

1.1. Solution

Let x∈G be odd, let H be the set { y ∈ G | y is even }, and let Hx be the coset { yx | x ∈ H }. Then there is a one-to-one correspondence between H and Hx, so |H| = |Hx|. Note that every permutation in Hx is odd. The converse also holds: if z is an odd permutation in G, then zx^-1 is even and thus in H, and so zx^-1x = z is in Hx. It follows that every permutation in G is in exactly one of H and Hx, so |G| = |H| + |Hx| = 2|H| is even.

2. Damaging a group

Let G be a group. Consider the algebra G^* obtained by replacing the multiplication operation in G with x*y = xy^-1 and G^** obtained by replacing the multiplication operator in G with x**y = x^-1y^-1 (where in each case multiplication and inverses are done using the original operation in G).

1. Prove or disprove: For any group G, G^* is a semigroup.

2. Prove or disprove: For any group G, G^** is a semigroup.

2.1. Solution

Neither is a semigroup, because neither satisfies associativity:

1. (x*y)*z = (x*y)z^-1 = xy^-1z^-1 but x*(y*z) = x*(yz^-1) = x(yz^-1)^-1 = xzy^-1.

2. (x**y)**z = (x**y)^-1z^-1 = (x^-1y^-1)^-1z^-1 = yxz^-1 but x**(y**z) = x^-1(y**z)^-1 = x^-1(y^-1z^-1)^-1 = x^-1zy.

3. Whirling polygons

Show that D_n has a subgroup of size m if and only if m divides 2n.

3.1. Solution

The only if part is just Lagrange's theorem.

For the if part, first recall that D_n is generated by a flip f and a rotation r, with rⁿ = f² = e and fr = r^-1f. So in particular the subgroup generated by r is isomorphic to ℤ_n. It follows that D_n has a subgroup (generated by r^n/m) for each m that divides n.

Suppose now that m does not divide n but does divide 2n. Then m = 2k where k∣n (we can only fit one extra 2 in 2n). Consider the subgroup of D_n generated by f and r^n/k; this has exactly k elements of the form r^an/k and k of the form r^an/kf, for a total of 2k=m elements.

4. Cocosets

Given a group G with subgroups H and K, define HK = { hk | h ∈ H, k ∈ K }. Show that HK is a subgroup of G if and only if HK = KH.

4.1. Solution

Let's do the if direction first. Let hk ∈ HK and consider (hk)^-1 = k^-1h^-1 ∈ KH = HK. Now consider rs ∈ HK; we wish to show hkrs is also in HK. But since kr ∈ KH = HK there exists some uv ∈ HK such that kr = uv. Rewrite hkrs = huvs = (hu)(vs) ∈ HK.

For only if, observe that if HK is a subgroup, then for each x ∈ G we have x ∈ HK if and only if x^-1 ∈ HK (the only if part is just closure of H under inverses applied to (x^-1)^-1 = x). But then HK = { hk | h ∈ H, k ∈ K } = { (hk)^-1 | h ∈ H, k ∈ K } = { k^-1h^-1 | h ∈ H, k ∈ K } = { kh | k ∈ K, h ∈ H } = KH.

5. Rational quotients

Let ℚ be the additive group of the rationals, i.e. the group whose elements are numbers of the form n/m for integers n and m ≠ 0 and whose operation is the usual addition operation for fractions, and let ℤ be the additive group of the integers, which we will treat as equal to the subgroup of the rationals generated by 1 = 1/1. Prove or disprove: ℤ is isomorphic to a subgroup of ℚ/ℤ.

5.1. Solution

Let f:ℤ→ℚ/ℤ be a homomorphism. We will show that f is not injective: there exist n and m in ℤ such that f(n) = f(m). Since any isomorphism between ℤ and a subgroup of ℚ/ℤ would give an injective f, this will show that no such isomorphism exists.

Let f(1) = a/b. Then f(b+1) = (b+1) f(1) = (b+1)(a/b) = a/b + a. But a ∈ ℤ, so a/b + a is in the same coset of ℤ as a/b. It follows that f(1) = f(b+1) and f is not injective.

Comment: The quotient group ℚ/ℤ is generally known as the rationals mod 1, since a natural choice of representatives for the various cosets is the rationals x with 0 ≤ x < 1, and addition yields the remainder after subtracting out any ones. This is similar to the construction of ℤ_m ≈ ℤ/mℤ.