Question

Let g1, g2,... gr be representatives of conjugacy classes in G. Let g; commute with 9j for all i, j. We have to prove that G is abelian. It is enough to prove that |G|= r. Consider Cl(g;). Without loss of generality we assume that _g₁ = e. Let Cl(g;) denote the conjugacy class of gi. Now, Cl(gi)| = |G|/|Cc(gi)|'| where CG(gi) denote the centralizer of gi. Since each g; commute with gj, we have gk Є CG(gi), for every i, k. So |Cl(gi)| ≤ |G|/r. Now we write the class equation:

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|G| = ∑ Σi=1|Cl(gi)| ⇒ |G| ≤ 1 + "=¹|G| ⇒ |G|≤ r.
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But this proves G =r, and we are done.

Answer 1

The original assumption that G is not abelian must be false. This means that G must be abelian.

The proof that G is abelian if g1, g2, ..., gr are representatives of conjugacy classes in G, and gi commutes with gj for all i and j:

We will prove this by contradiction. Assume that G is not abelian.

This means that there exist elements a and b in G such that ab ≠ ba.

Let H be the subgroup of G generated by a and b. Since H is a subgroup of G, it is also closed under conjugation.

This means that for all h in H and g in G, ghg⁻¹ is also in H.

Let ha and hb be the conjugates of a and b in H, respectively. Since H is closed under conjugation, ha and hb are also in H. We know that ab ≠ ba, so hahb ≠ hbhb.

Let C(hb) be the centralizer of hb in H. This is the subgroup of H consisting of all elements that commute with hb.

Since hahb ≠ hbhb, ha ∉ C(hb). This means that there exists an element h in H such that hah⁻¹ ≠ hb.

Let k = hah⁻¹. Since H is closed under conjugation, k is also in H. We know that k ≠ e, since hah⁻¹ ≠ hb.

We claim that k commutes with all elements of H. To see this, let h be any element of H. Then we have:

khk⁻¹ = (hah⁻¹)h(hah⁻¹)

= hah⁻¹hhah⁻¹

= ha²h⁻¹

= ha(hb)h⁻¹

= (hah⁻¹)(hb)

= hb

This shows that khk⁻¹ = hb for all h in H, so k commutes with all elements of H.

ince k commutes with all elements of H, it must also commute with all elements of the centralizer of hb in H.

This means that k ∈ C(hb). However, we already saw that ha ∉ C(hb), so we have a contradiction.

Therefore, our original assumption that G is not abelian must be false. This means that G must be abelian.