# Showing that all strings representing a group element end the same way

I'm currently working with a group given by the presentation $$\bigl<R,T\,\bigm|\,T^2=(RT)^3=1\bigr>,$$ and I'm trying to prove a result applying to any element with defining string ending in T: every string equivalent to it that

a) contains no inverse symbols and

b) does not contain T more than once in a row

also either ends in T or has a suffix equivalent to T.

My thought is that any string equivalent to the given string is made by multiplying by conjugates of $RTRTRT$ and $TT$, but in order for the resulting string not to contain any inverse symbols, if you multiply by $ARTRTRTA^{-1}$ for an element $A$ then the string you're multiplying by must start with $A$ so that the inverse cancels it out. But this will be the same as if you insert the $RTRTRT$ after the initial string $A$. Also, since one can't have two $T$'s in a row, inserting $TT$ is not allowed, and any pair of $T$'s must be canceled upon appearing in the string.

So now, if the only way equivalent strings can be made is by inserting $RTRTRT$ somewhere in the string and canceling $T$'s, it would appear that all of the strings equivalent to one ending with T also have a suffix equivalent to T.

Does this reasoning work? If not, how can I fix it?

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Can you not try and re-formulate your question using free products, and the normal-form theorem which goes along with it? $R\mapsto RT^{-1}$ gives you that your group is just $C_2\ast C_3$. I apologise for not being able to think about this properly - I am holding a sleeping baby so am somewhat handicapped! –  user1729 Sep 4 '11 at 17:20
I don't really understand how that helps in this problem; I'm looking at all the strings that represent a given element, not just the canonical form. What am I missing? –  William D. Sep 4 '11 at 19:39
The main difficulty I find with your argument is that you can have a "hidden" inverse symbol. For example, since $R^{-1} = TT^{-1}R^{-1}= T(RT)^{-1} = T(RT)^2$, you can replace any instance of $R^{-1}$ with $T(RT)^2$; and of course any instance of $T^{-1}$ can be replaced with $T$. So in fact, any word whatsoever can be written with no inverse symbols, so conjugation can be effected without the introduction of any inverse symbols. I haven't thought it through to know if this invalidates your chain of reasoning, but I don't see it being taken into account. –  Arturo Magidin Sep 4 '11 at 21:49
Well, the only thing that T's hidden inverse amounts to is that one is able to insert $(TR)^3$ instead of $(RT)^3$, because any other T's have to be canceled to avoid having two in a row. I'm working on the case of conjugation by R, but I don't think it will be much of a problem. –  William D. Sep 5 '11 at 0:52
Well, and R's hidden inverse shouldn't be a problem either. You can derive it by taking the fact that $RTRTRT=I$ (the identity) and just multiplying both sides by $R^{-1}$ gives the hidden inverse. So conjugating $RTRTRT$ by $R$ is essentially inserting $RTRTRT$ after the $R$ in $RTRTRT$. So it looks like the problem of conjugation is already solved because it amounts to inserting $RTRTRT$ within itself. Is this valid? –  William D. Sep 5 '11 at 3:09