# A Level P6 exam

Jummeh
I was revising for my ALevel P6 exam the other day and I came up against a very tough question...

G is a multiplicative froup with identity element e, and a is a fixed element of G for which axa = x^-1 for all elements x belonging to G. Prove that...

i/ a = a^-1

ii/ ax = (ax)^-1 for all x belong to G //cant find the little symbol

iii/ x = x^-1 for all x belong to G

iv/ xy = yx for all x, y belong to G
I can do parts i and ii

axa = x^-1, let x = e
a^2 = e
a = a^-1

for part ii

axa = x^-1
axax = e
axax(ax)^-1 = (ax)^-1 //ax and (ax)^-1 will cancel
ax = (ax)^-1

but for part iii and iv I seem do always go around in circles or i end up with something obvious like x = x. Unfortunately there aren't any answers to this question.

ps: I hope this was the correct place to post this, i had a look in the homework help section but it was mostly about physics. :s
And i have already taken the exam now but the question is kinda bugging me.

Thanks in advance for any help

Jim

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## Answers and Replies

Staff Emeritus
Gold Member
The result of part two is the key to solving answer three. Maybe if I slightly restate what you've proven, the way to go might become apparent:

Theorem: for all y, (ay) = (ay)^-1

Goal: prove x=x^-1 for all x

And result 4 follows from a few applications of rule 3 and the identity:
(xy)^-1 = (y^-1)(x^-1)