10.4 Some applications of Lagrange’s theorem

Lagrange’s Theorem gives most information about a group when the order of the group has relatively few factors, as then it puts more restrictions on possible orders of subgroups and elements.

Let’s consider the extreme case, when the order of the group is a prime p, and so the only factors are 1 and p.

Theorem 10.21:

Let p be a prime and G a group with |G|=p. Then

  1. G is cyclic.

  2. Every element of G except the identity has order p and generates G.

  3. The only subgroups of G are the trivial subgroup \{e\} and G itself.

Proof.

Let x\in G with x\neq e. Then \operatorname{ord}(x) divides |G|=p by Corollary 10.20, and \operatorname{ord}(x)\neq1 since x\neq e. So \operatorname{ord}(x)=p. So the cyclic subgroup \langle x\rangle generated by x has order p, and so must be the whole of G. This proves a.) and b.).

Let H\leq G. Then by Theorem 10.15 |H| divides |G|=p, and so |H|=1, in which case H is the trivial subgroup \{e\}, or |H|=p, in which case H=G.

Remark:
In particular this shows that if p is prime then all groups of order p are isomorphic.
Corollary 10.22:

If p is prime and P and Q are two subgroups of a group G with |P|=p=|Q|, then either P=Q or P\cap Q=\{e\}.

Proof.

If P\cap Q\neq\{e\} then choose x\in P\cap Q with x\neq e. By the previous theorem, x generates both P and Q, so P=\langle x\rangle=Q.

Now some other simple general consequences of Lagrange’s Theorem.

Proposition 10.23:

Let G be a group and H,K two finite subgroups of G with |H|=m, |K|=n and \gcd(m,n)=1. Then H\cap K=\{e\}.

Proof.

Recall that the intersection of two subgroups is itself a subgroup, so that I=H\cap K is a subgroup both of H and of K. Since it’s a subgroup of H, Lagrange’s Theorem implies that |I| divides m=|H|. But similarly |I| divides n=|K|. So since \gcd(m,n)=1, |I|=1 and so I=\{e\}.

Theorem 10.24:

Let G be a group with |G|=4. Then either G is cyclic or G is isomorphic to the Klein 4-group C_2\times C_2. In particular there are just two non-isomorphic groups of order 4, both abelian.

Proof.

By Corollary 10.20 the order of any element of G divides 4, and so must be 1, 2 or 4.

If G has an element of order 4 then it is cyclic.

If not, it must have one element (the identity e) of order 1 and three elements a,b,c of order 2. So a^{-1}=a, b^{-1}=b and c^{-1}=c.

Consider which element is ab. If ab=e then b=a^{-1}, which is false, since a^{-1}=a. If ab=a then b=e, which is also false. If ab=b then a=e, which is also false. So ab=c.

Similarly ba=c, ac=b=ca and bc=a=cb, and G is isomorphic to the Klein 4-group.

We’ll finish this section with some other results about groups of small order that we won’t prove. These are easier to prove with a bit more theory, which are all proved in the third year Group Theory unit.

Theorem 10.25:

Let p be an odd prime. Then every group of order 2p is either cyclic or isomorphic to the dihedral group D_{2p}.

Theorem 10.26:

Let p be a prime. Every group of order p^2 is either cyclic or isomorphic to C_P\times C_p (and so in particular is abelian).

However there are non-abelian groups of order p^3 for every prime p. The dihedral group D_8 is one example for p=2.

Theorem 10.27:

There are five groups of order 8 up to isomorphism. Three, C_8, C_4\times C_2 and C_2\times C_2\times C_2, are abelian, and two, the dihedral group D_8 and another group Q_8 called the quaternion group are non-abelian.

The first few orders not dealt with by the general theorems above are 12, 15, 16 and 18. It turns out that are five non-isomorphic groups of order 12, every group of order 15 is cyclic, there are fourteen non-isomorphic groups of order 16, and five of order 18.

The number of non-isomorphic groups of order 2^n grows very quickly with n. There are 49,487,365,422 (nearly fifty billion) non-isomorphic groups of order 1024=2^{10}.