 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ia/paper-3/1d/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Suppose

G / Z (G)

is generated by

g Z (G)

. Every element of

G

can then be written

g^{a} z

with

z \in Z (G)

. Hence

(g^{a} z_{1}) (g^{b} z_{2}) = g^{a + b} z_{1} z_{2} = (g^{b} z_{2}) (g^{a} z_{1}) .

G

to be abelian.

∣ H / Z (H) ∣ = 7

, the quotient has prime order and is therefore cyclic. The result just proved would make

H

abelian, so

Z (H) = H

and the quotient would have order one, a contradiction. Therefore

no such group H exists .

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