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Is this true?

If $G$ is a group of size $n$, and $X$ is a non-empty subset of $G$ then $X^n$ is a subgroup of $G$?

By $X^n$ I mean the set of all products of length $n$ from $X$.

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EDIT: please ignore this answer, which was based on not understanding the problem.

There's a standard way to check whether something is a subgroup.

Is it non-empty? Sure.

Is it closed under the operation? $$(a_1a_2\cdots a_n)(b_1b_2\cdots b_n)=(a_1a_2)(a_3a_4)\cdots(b_{n-1}b_n)$$ so that works.

Does it have the inverse of each of its elements? Sure, the inverse of a product of $n$ things is the product of the $n$ inverses (in the opposite order).

All done.

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    $\begingroup$ I think you misunderstood the question: the question is about the product of length $n$ of elements of a given subset of a finite group, not the entire group. $\endgroup$
    – tomasz
    Oct 13, 2012 at 11:29
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    $\begingroup$ Besides, as you have interpreted it, the solution is quite trivial: for any $g\in G$ we have $g=g\cdot e^{n-1}$, so the set is just the entire $G$, hence a group. $\endgroup$
    – tomasz
    Oct 13, 2012 at 11:32
  • $\begingroup$ @tomasz, yes, I thought it was too easy. $\endgroup$ Oct 13, 2012 at 12:13
  • $\begingroup$ The argument is wrong. You implicitly used the fact that a product of two elements of $X$ and the inverse of an element of $X$ lies again in $X$. Thats it what you want to proove so you have a circle argument. $\endgroup$ Sep 14, 2014 at 11:39

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