1. ## Group theory

Let G be a finite group of order n with identity element e. If $\displaystyle a_1, ..., a_n$ are n elements of G, not necessarily distinct, prove that there are integers p and q with $\displaystyle 1 \leq p \leq q \leq n$ such that $\displaystyle a_p a_{p+1} ... a_q = e$.
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I have not got very far in proving this apart from a few basic cases. If any of the elements on the list are the identity, it is trivial. This means that you can assume it is a list of length n which contains at most (n-1) distinct elements of G. So at least one element occurs twice. I took some cases after this, but every case is difficult and seems to lead to taking more cases . I'm sure there is a simpler way... can anyone help?

2. Originally Posted by Capillarian
Let G be a finite group of order n with identity element e. If $\displaystyle a_1, ..., a_n$ are n elements of G, not necessarily distinct, prove that there are integers p and q with $\displaystyle 1 \leq p \leq q \leq n$ such that $\displaystyle a_p a_{p+1} ... a_q = e$.
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I have not got very far in proving this apart from a few basic cases. If any of the elements on the list are the identity, it is trivial. This means that you can assume it is a list of length n which contains at most (n-1) distinct elements of G. So at least one element occurs twice. I took some cases after this, but every case is difficult and seems to lead to taking more cases . I'm sure there is a simpler way... can anyone help?
Consider the products $\displaystyle a_1a_2\cdots a_k$ for k=1,2,...,n. Either one of them is the identity (in which case problem solved) or two of them are the same, say $\displaystyle a_1a_2\cdots a_{p-1} = a_1a_2\cdots a_q$, ... .