Proving $U=U$ is derivable from any set of relators. This is Exercise 1.1.1 of Magnus et al's book on combinatorial group theory.
The details:
Let $G=\langle X\mid R\rangle$ for sets $X, R$.

Definition 1: The empty word $\varepsilon$ and the words $\{aa^{-1}\mid a\in X\}\cup \{b^{-1}b\mid b\in X\}$ are called trivial relators.
Definition 2: Suppose $\{P, Q, \dots\}\subseteq R$. Then $W$ is derivable from $\{P, Q, \dots\}$ if the following operations, applied a finite number of times, turn $W$ into $\varepsilon$:
(i) Inserting one of $P, P^{-1}, Q, Q^{-1}, \dots$ or one of the trivial relators between two consecutive symbols of $W$, or before $W$, or after $W$.
(ii) Deleting one of $P, P^{-1}, Q, Q^{-1}, \dots$ or one of the trivial relators, if either forms a subword of $W$.
Definition 3: We say the equation $W=V$ is derivable from the relators $P_1=P_2, Q_1=Q_2, \dots$ (of $G$) exactly when $WV^{-1}$ is derivable from the relators $P_1P_2^{-1}, Q_1Q_2^{-1}, \dots (\in R)$.

The Question:

Show (A) by induction on the length of the word $U$ that $UU^{-1}$ is derivable from the trivial relators, and hence (B) that the relation $U=U$ is derivable from any set of relators.
[Hint: If $U=Va^\eta$, where $\eta=\pm 1$, then $UU^{-1}=Va^\eta a^{-\eta}V^{-1}$.]

My Attempt:
That (B) follows from (A) by definition, is obvious. Thus it suffices to prove (A).
Let $U$ be a word in $X$.
We proceed by induction on the length of $U$.
Suppose $\lvert U\rvert=0$. Then $U=\varepsilon$. One applies (i) and (ii) zero times to get the empty word from $UU^{-1}=\varepsilon\varepsilon=\varepsilon$. Thus $UU^{-1}$ is derivable from the trivial relators.
Assume that for $\lvert U\rvert=u$, we have that $UU^{-1}$ is derivable from the trivial relators.
Consider when $\lvert U\rvert=u+1$. Then $U=Va^\eta$ for some $a\in X$, $\eta=\pm 1$, $V$ a word in $X$ s.t. $\lvert V\rvert=u$, so that $UU^{-1}=Va^\eta a^{-\eta}V^{-1}$. The word $a^\eta a^{-\eta}$ is trivial, so apply (ii) once to get $VV^{-1}$, which is itself derivable from the trivial relators by the induction hypothesis.
Hence $UU^{-1}$ is derivable from the trivial relators. $\square$


Is this proof rigourous enough?

Thoughts:
The step . . .

"The word $a^\eta a^{-\eta}$ is trivial, so apply (ii) once to get $VV^{-1}$, which is itself derivable from the trivial relators by the induction hypothesis.
"Hence $UU^{-1}$ is derivable from the trivial relators."

. . . could do with some justification, I think.

Please help :)
 A: I think there are two improvements you can make. Once is minor, the other major.


*

*In the step you ask about, I think it would be good if you wrote "$UU^{-1}=Va^{\eta}a^{-\eta}V^{-1}=VV^{-1}$, by (ii), so the result holds by induction". That is, I think you need to write the actual maths/use notation here to make it clear what is going on.

*Your base case is wrong. You actually need to justify that $\epsilon\epsilon=\epsilon$. This holds because $\epsilon$ is a trivial relator so you can removed one of them (the left or right one - you choose!) via (ii). So one application of (ii), not zero.
As a side-note, I still remember my PhD supervisor saying that these exercises were an amazing achievement. I am not sure how many of them I have actually done though...(I think my supervisor might have done them all).
[Also, try to avoid the words "clearly" and "obvious". If it is clear then you don't need to say so. If you feel like something is missing from your explanation then neither of these words will fill that gap! So instead of
"That (B) follows from (A) by definition, is obvious."
write
"By Definition 3, (B) follows from (A)."]
