Group Theory: Math30038, Sheet 3

GCS

Revised 31-x-2003

The course web site is available via http://www.bath.ac.uk/∼masgcs/

1. Let ϕ denote the Euler ϕ-function. Prove that for every integer n we

have

∑ ϕ(d) = n

where the sum is taken over all natural numbers d which divide n. Solution: In a cyclic group of order n, there are exactly ϕ(d) elements of order d where d is any divisor of n.

2. Suppose that G is a finite abelian group. Suppose that p is a prime number which divides |G|. Prove that there is an element g ∈ G such that o(g) = p. Hint: multiply together all the cyclic subgroups of G. Solution: Follow the hint. If each element of G has order coprime to p, then it follows from the formula for the size of a product of subgroups that G has size coprime with p. This is not the case so there is an element x ∈ G of order pm. Now y = x^{m} has order p.

3. Suppose that G is a group and that K, L are both normal subgroups with the property that K ∩ L = 1 (i.e. K and L intersect to form the trivial subgroup consisting of the identity element). Prove that every element of K commutes with every element of L. Hint: consider elements of the form k^{−1}l^{−1}kl where k ∈ K and l ∈ L. Solution: k^{−1}L^{k} = L and l^{−1}Kl = K by normality so k^{−1}l^{−1}k ∈ L and l^{−1}kl ∈ K. Therefore k^{−1}l^{−1}kl ∈ K ∩ L = {1}. Thus kl = lk wherever k ∈ K and l ∈ L.

4. Suppose that G is a group and that H is a subgroup of G of finite index. Suppose that K is also a subgroup of G. Prove that |K : H ∩ K| ≤ |G : H|. What can you say if this inequality is an equality? Solution: Let T be a right transversal for H ∩ K in K, so if t, t′ ∈ T are distinct, then t, t′ ∈ K but tt^{′−1} ∉ H ∩ K. Therefore tt^{′−1} ∉ H. It follows that Ht, Ht′ are distinct cosets. The inequality is established. We claim that equality holds if and only if HK = G. We prove this as follows. If HK = G then it is possible to choose a right transversal S for H in G consisting of elements of K. Now if s, s′ are distinct elements of S then ss^{′−1} ∉ H and so ss^{′−1} ∉ H ∩ K. Therefore |G : H| ≤ |K : H ∩ K|. However, the reverse inequality was established earlier Thus |G : H| = |K : H ∩ K|. Next suppose insteab that |G : H| = |K : H ∩ K|. The elements u of right transversal U for H ∩ K in K give rise to distinct right cosets Hu (why?), but by assumption these are all the right cosets of H in G, thus HU = G and therefore HK = G.

5. Let G be a group. Suppose that H ≤ G and that |G : H| = 2. Prove that H ⊴ G. Can one derive the same conclusion when 2 is replaced by 3? Solution: Suppose that g ∈ G. If g ∈ H, then gH = H = Hg. On the other hand if g ∉ G, then gH ≠ H, but there are only two left cosets of H in G so gH = G−H (i.e. the set of elemenst of G which are not elements of H). Similarly Hg = G − H so gH = Hg. Thus H is a normal subgroup of G. The argument will not work when 2 is replaced by 3, for we may let G = S_{3}, and put H = ⟨(1, 2)⟩. Now H has order 2 and therefore index 3 in G. Moreover (2, 3)H = {(2, 3), (1, 2, 3)} but H(2, 3) = {(2, 3), (1, 3, 2)}.