 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2005/paper-2/8/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

A left coset is

g H = {g h : h \in H}

. The cosets partition a finite group into equally sized sets of

∣ H ∣

elements, so Lagrange's theorem gives

∣ G ∣ = [G : H] ∣ H ∣

; in particular subgroup order divides group order.

Here

A = - I

C = - B

and

B^{2} = I

. The Cayley table, with row element multiplied on the left of the column element, is

$\cdot$	$I$	$A$	$B$	$C$
$I$	$I$	$A$	$B$	$C$
$A$	$A$	$I$	$C$	$B$
$B$	$B$	$C$	$I$	$A$
$C$	$C$	$B$	$A$	$I$

The table proves closure,

I

is the identity and every element is its own inverse. Associativity is inherited from matrix multiplication, so this is the Klein four-group.

Finally, if every nonidentity element has order two, every element satisfies

h^{- 1} = h

, including the identity. Then

ab = (ab)^{- 1} = b^{- 1} a^{- 1} = ba

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