## Maths.ox.ac.uk

Finite groups and their characters (Paper B2b)
Hilary Term 2007: Sheet 0: Specimen answers
Remind yourself about Lagrange’s Theorem and its proof.

Show that Alt (4) has no subgroup of order 6.

See any textbook of algebra or group theory. Indeed, it is a good idea to see several and compare
their treatments of Lagrange’s Theorem.

Let G := Alt (4) and suppose that G had a subgroup H of order 6.

|G : H| = 2, so H would be normal in G (see Qn 5, or note that the two left cosets of H are H andG \ H, as are its two right cosets, so left and right cosets coincide, and this is one of the ways to definea normal subgroup). Therefore there would be a non-trivial homomorphism φ : G → Z2. But Alt (4) isgenerated by its 3-cycles, and since these have odd order they must map to the identity in Z2. Henceany homomorphism G → Z2 is trivial. Thus Alt (4) cannot have a subgroup of order 6.

Method 2. Suppose that Alt (4) had a subgroup H of order 6. Then |Alt (4) : H| = 2, and so if t is
any 3-cycle in Alt (4) then two of e, t, t2 must lie in the same right coset of H, whence either t ∈ H ort−1 ∈ H, and therefore in fact t ∈ H. Thus H contains all the 3-cycles. But there are 8 of these. Thiscontradiction shows that Alt (4) cannot have a subgroup of order 6.

(ii) that Zn has a subgroup of order m if and only if m n ;
(iii) that if m n then there is only one subgroup of order m in Zn.

See any textbook of algebra or group theory. Indeed, it is a good idea to see several and compare
Find all the groups of order < 8 up to isomorphism.

8 See, for example, W Ledermann, Introduction to group theory,
Longman 1973, § 14. The groups of order n are: Zn for all relevant n and, in addition, Z2 × Z2 for n = 4(this group is often called the four-group and denoted V4); and Sym (3) (which is isomorphic to D3) forn = 6.

Extending the question a little, There are five groups of order 8, two of order 9, two of order 10, one
Let G be any group, H a subgroup of it. The core of H in G is defined by
Now suppose that m := |G : H| < ∞ and that coreG(H) = {e}. (Then H is said to be core-free in
and so g−1coreG(H)g = coreG(H). Thus coreG(H)
Consider the permutation representation ρ : G → Sym (m) produced by the action of G on the set
of right cosets of H by right multiplication. We find that
Ker ρ = {g ∈ G | Hxg = Hx for all x ∈ G} = {g ∈ G | xgx−1 ∈ H for all x ∈ G}
= {g ∈ G | g ∈ x−1Hx for all x ∈ G} =
that is, Ker ρ = coreG(H) [compare Neumann, Stoy & Thompson, Groups and Geometry, p. 64]. Thefact that coreG(H) = {e} therefore means that ρ is an injective homomorphism (so an isomorphism toits image), hence |G| = |Im (ρ)|. Since Im (ρ)
Sym (m), Lagrange’s Theorem tells us that |G| m!, as
required. Notice that, in particular, G is finite and of order at most m!.

Show that in any group a subgroup of index 2 is normal. Show also that in a finite group of odd
order a subgroup of index 3 is normal.

Let H be a subgroup of index 2 in the group G. One method for proving that H
in the discussion of Qn 1 above. Alternatively, as in Qn 4, consider the permutation representationρ : G → Sym (2) produced by the action of G on the set of right cosets of H by right multiplication.

Its image is a non-trivial subgroup of Sym (2), therefore must have order 2, and so |G : Ker ρ| = 2. ButKer ρ
H. It follows that Ker ρ = H and therefore H
This alternative method works for the second part. Let G be a finite group of odd order, and H a
subgroup with |G : H| = 3. Consider the permutation representation ρ : G → Sym (3) produced by theaction of G on the set of right cosets of H by right multiplication. Its image is a non-trivial subgroupof Sym (3) of odd order, and therefore must have order 3. Therefore |G : Ker ρ| = 3. But Ker ρ
For (i) we need first to prove that Z(G) is a subgroup of G. Well, certainly e ∈ Z(G); if z ∈ Z(G)
then zg = gz for all g ∈ G so gz−1 = z−1g for all g ∈ G, which means that z−1 ∈ Z(G); and ifz1, z2 ∈ Z(G) then for any g ∈ G we have z1z2g = z1gz2 = gz1z2, so z1z2 ∈ Z(G). Hence Z(G)
Now for any x ∈ G, z ∈ Z(G) we have x−1zx = x−1xz = z, and therefore in fact Z(G)
3 and let g ∈ Sym (n) \ {e}. If g has more than one cycle (including cycles of
length 1) choose α, β in different cycles of g and let h be the transposition (α β). Then αgh = αg,whereas αhg = βg, and therefore gh = hg, so g /
∈ Z(Sym (n)). Alternatively, g has just one cycle, so
g = (α1 α2 . . . αn). Then let h be the transposition (α1 α2). Again we find that gh = hg (this is whereit is relevant that n
∈ Z(Sym (n)). Therefore Z(Sym (n)) = {e}.

4 and let g ∈ Alt (n) \ {e}. If g has more than one cycle of length at least 2
suppose that g = (α1 α2 . . . αk)(β1 β2 . . . βm) · · · , where k
be the 3-cycle (α1 α2 β1). Then αgh = αh = β
i = 1 if k = 2. In any event, gh = hg, so g /
∈ Z(Alt (n)). Alternatively, g has just one non-trivial
n. Note that k must be odd since g is an even permutation. If
5 then take h to be the 3-cycle (α1 α2 α3), while if k = 3 then take h to be the double transposition
(α1 α2)(α3 β) where β is any element other than α1, α2, or α3. This is possible since n
∈ Z(Alt (n)). Therefore Z(Alt (n)) = {e}.

Source: http://www.maths.ox.ac.uk/files/imported/current-students/undergraduates/lecture-material/B/groups/pdf/groups-sol0.pdf

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