Subgroups, cosets, Lagrange, cyclic groups

Lagrange and its partial converses

|H| divides |G|, and o(g) divides |G|, so $g^{|G|}$ = e.

Statement	True?
d | |G| ⇒ G has a subgroup of order d	✗ — $A_{4}$ has order 12, no subgroup of order 6
p prime, p | |G| ⇒ G has an element of order p	✓ Cauchy
pᵏ | |G| ⇒ G has a subgroup of order pᵏ	✓ Sylow
G abelian and d | |G| ⇒ subgroup of order d	✓
G cyclic ⇒ exactly one subgroup of each order dividing |G|	✓

Cyclic groups

$\mathbb{Z}_{n}$ has $\varphi(d)$ elements of order d for each d | n, and exactly one subgroup of each such order. |Aut$(\mathbb{Z}_{n})$| $= \varphi(n)$.

Union of proper subgroups: a finite group is the union of its proper subgroups iff it is not cyclic — a generator lies in no proper subgroup, and conversely every element generates a proper one. No group over any field is the union of two proper subgroups.

Order arithmetic

• o(ab) = o(a)o(b) when a, b commute and gcd(o(a), o(b)) = 1. Without commuting it can be infinite: in GL$_{2}$ two elements of order 2 can multiply to infinite order.
• Index 2 ⇒ normal. Index p (smallest prime dividing |G|) ⇒ normal.
• |G| = p ⇒ cyclic. |G| $= p^{2} \Rightarrow$ abelian $(\mathbb{Z}_{p^{2}}$ or $\mathbb{Z}_p \times \mathbb{Z}_p)$, not necessarily cyclic.
• |G| = pq with p < q and p ∤ q − 1 ⇒ cyclic.

Key takeaways

• Converse of Lagrange fails $(A_{4}$, order 6); Cauchy and Sylow are the true partial converses.
• Non-cyclic ⇔ union of proper subgroups. $- p^{2} \Rightarrow$ abelian, but not necessarily cyclic.