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Group Theory

2 Miscellaneous

Proposition 2.1 (Lagrange’s Theorem). If \(H\) is a subgroup of \(G\) with \(G\) finite, then \(|H|\) divides \(|G|\).

Proof. \(G\) acts transitively on each coset in \(G/H\), which has size \(H\). □

In particular \(|G/H| = |G|/|H|\) if \(G\) finite. We call \(|G/H|:=[G:H]\) the index of \(H\) in \(G\). Note that index is multiplicative.

Proposition 2.2. If \(H,N \triangleleft G\), \(H\cap N = {1}\), \(HN = G\), then \(G \cong N \times H\).

Proof. We have \(G/H \cong N\) and \(G/N \cong H\) so the product of the projections gives a map \(G \to N \times H\), which is bijective as \(HN = G\) and \(H\cap N = {1}\). □

Lemma 2.3. If \(H,N\subset G\) are subgroups, and either \(H,N\) is normal, then \(HN\) is a subgroup. If \(H,N\) are normal, then \(HN\) is normal.

Proof. For the first statement look at the second isomorphism theorem, and for the second statement just note \(HN/N\) is normal in \(G/N\) since \(H\) is normal in \(G\). □

Lemma 2.4. If \(H,N\subset G\) are subgroups which commute with each other, then \(HN\) is a subgroup and \(|HN| = \frac {|H||N|}{|H\cap |N|}\).

Proof. \(h_1n_1h_2n_2 = h_1h_2n_1n_2\) shows that \(HN\) is a subgroup. We have a homomorphism from \(H\times N\) to \(HN\) sending \((h,n) \mapsto hn\) which is surjective, with kernel in bijection with pairs \((h,n)\) such that \(hn^{-1}=1\). □

Proposition 2.5 (Class Equation). \(|G| = |Z(G)| + \sum _{|Cl|>1}|Cl|\).

Proof. This is just the statement that the conjugation action partitions \(G\) into orbits. □

Theorem 2.6 (Classification of finitely generated abelian groups). Every finitely generated abelian group is uniquely of the form \(\ZZ ^r\oplus \bigoplus _1^k\ZZ /n_i\ZZ \) with \(n_i|n_{i+1}\).

Proof. This is immediate after realizing that abelian groups are \(\ZZ \)-modules. □

Theorem 2.7 (Cauchy’s Theorem). If \(p||G|\), then there is an element of order \(p\) in \(G\).

Proof. WLOG we may assume that \(p\nmid |Z(G)|\) as otherwise either \(G\) is abelian in which case Theorem 2.6 makes this obvious, or we can induct, and inspect \(Z(G)\) instead. Now by looking at the class equation, we must have some nontrivial conjugacy class with \(p\) not dividing its size, so we can induct by looking at the corresponding centralizer. □

Lemma 2.8. If \(H\) is a proper subgroup of \(G\) with \(|G| \nmid [G:H]!\), then \(G\) is not simple.

Proof. \(G\) acts on \(G/H\) via left multiplication, this action is given by a homomorphism to \(S_{G/H}\), but by our assumption this must have a nontrivial kernel, which is proper as the action of \(G\) is nontrivial. □

In particular, infinite simple groups cannot have proper subgroups of finite index.

Lemma 2.9. \([G:Z(G)]\) cannot be prime.

Proof. If the center is not everything, then the centralizer of an element not in the center is a subgroup strictly between \(G\) and \(Z(G)\). □