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G is a group and S is a nonempty finite subset which is closed under multiplication. Show that S is a subgroup of G.

My Attempt

Let S be equal to $\{a_1,a_2.....a_n$} and let it be ordered such that $ a_1a_2.....a_n = a_1$. Multiplying both sides with $a^{-1})$ we get $a_2.....a_n = e$.

Thus e∈S.

$(a_2.....a_n)^{-1}$ = e

$a_n^{-1}.....a_2^{-1}$ = e

We can show that any $a_k^{-1}$ between $a_n^{-1}$ and $a_2^{-1}$ is an element of S by multiplying by $a_n^{}$, $a_{n-1}$.... until we reach $a_{k}$. We can do the same thing from the right starting with $a_2$.

Here's where I'm stuck. I have proved that the inverses for all elements in S exist in S except for $a_1$. Any help on how to prove that would be appreciated.

Also, since I'm new to algebra, I don't know whether the method of multiplying on the left and the right until we reach $a_k$ is formally correct or not. So a more formal way of stating the same would also be really appreciated.

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    $\begingroup$ The statement you are proving is not true. For instance the subset $\{0,1,\dots\}$ of the group $(\mathbb Z,+)$ is not a subgroup but it is closed under the group operation (addition in this case). You need to assume that $G$ is a finite group! $\endgroup$ Jun 7, 2018 at 22:12
  • $\begingroup$ @TashiWalde Fixed, Thanks. $\endgroup$
    – LoneStar17
    Jun 7, 2018 at 22:14
  • $\begingroup$ @TashiWalde: Note that OP made $S$ finite while you suggested that the group has to be finite. It seems that making $S$ finite is sufficient but I haven't thought deeply about it. Certainly there is no finite subset of the integers except $\{0\}$ and the empty set that are closed under addition. The empty set is not closed because the empty sum results in $0$. $\endgroup$ Jun 7, 2018 at 22:41
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    $\begingroup$ @RossMillikan it is indeed enough to require S finite. See my answer. $\endgroup$ Jun 8, 2018 at 7:11

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You only need to show that $S$ is closed under inverses (then $1=aa^{-1}$ for any $a\in S$). So let $a\in S$. Consider the elements $a,a^2,a^3,\dots\in S$ since $S$ is finite, this sequence has to repeat at some point, lets say $a^n=a^m$ with $n<m$. But then $1=a^{m-n}$ (by multiplying both sides by $a^{-n}$), hence $a^{m-n-1}$ is the inverse of $a$ and lies in $S$ (note that if $m-n-1=0$ then $1=a^{m-n}=a$).

Regarding your proof: It is not at all obvious that there exists an ordering such that $a_1a_2\dots a_n=a_1$.

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  • $\begingroup$ I realized something after posting the question which i have edited in. Please have a look at it and see if it passes as an acceptable proof $\endgroup$
    – LoneStar17
    Jun 7, 2018 at 22:25
  • $\begingroup$ I don't understand where in your proof you conclude that $a_1^{-1}$ is in $S$. Also, you really need to justify why there is an ordering of the S such that $a_1\dots a_n=a_1$. $\endgroup$ Jun 7, 2018 at 22:32

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