1 – G is a group, ∀ a, b ∈ G, ab.
Pick u ∈ G.
a * = au⁻¹b.
(1) (G, *) is a group.
(2) What’s the inverse of u⁻¹ to *?

2 – Let G be a group a G ≠ {e}.
a subgroup N < G is called a maximal subgroup if N ≠ G, and if H < G and N ⊆ H ⊆ G, then H = N or H = G.
Each finite group contains a maximal subgroup.
{e} is not a subgroup of H (sub)1 is not a subgroup (denoted by ¬<) of H (sub)2 <…< H (sub)n, with no subgroups in between {e} and G, |G| < infinity.
If a finite group G ≠ {e} contains a unique maximal subgroup, then G is cyclic.
If H < G, then H is contained in the maximal subgroup of G.
H < H(sub)1, H < of H(sub)2.
Proof: Let N be the unique maximal subgroup of G.

3 – Theorems 4.5, 4.7, 4.8: [G : K] = |G| ÷ |K|

4 – G is nonabelian, and |G| = 2p, gcd (2, p) = 1.
(1) Prove ∃ a ∈ G, |a| = p.
∀ x ∈ G, |x|/2p ==> |x| = 2, p, or 2p, x ≠ e.
∃ x ∈ G, |x| = 2p. * ==> ∀ x ∈ G, |x| = 2p.
Suppose ∀ x ∈ G, x^2 = e.
Show G is abelian and try to get a contradiction.
(2) Let a ∈ G, |a| = p.
H = <a>, H = p.
Consider left cosets of H in G.
[G : H] = |G| ÷ |H| = 2p ÷ p = 2.
G = H ∪ gH

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