Subgroup of a finite group $H$ and $G$

Question

I regret to admit that this has been confusing me for much longer than I would like. I just can't wrap my head around some parts of this question and I'd like some guidance.

Let $G$ be a finite group, and let $S$ be a nonempty subset of $G$. Suppose $S$ is closed with respect to multiplication. Prove that $S$ is a subgroup of $G$. (HINT: It remains to prove that $S$ contains $e$ and is closed with respect to inverses. Let $S$ = {$a_1$ ... $a_n$}. If $a_i$ $∈$ $S$, consider the distinct elements $a_ia_1$, $a_ia_2$, $...$ $a_ia_n$

I think if I can understand one tiny segment then I can complete my proof. I know I need to show $e$ $∈$ $S$ and that $S$ is closed w.r.t inverses, but I can't get across a tiny step in proving $e$ $∈$ $S$.

The proof I'm trying to figure out goes something like:

Since $S$ is finite and closed under the group operation, then we have $a_1$ = $a_1a_k$ for some $a_k$ $∈$ $S$. Then from this we can find that $a_k$ = $e$.

My question is how we know that we have $a_1$ = $a_1a_k$ for some $a_k$ $∈$ $S$. I have no idea why this would be the case. I've also seen another post that used a map from $S$ $\rightarrow$ $S$ defined by left multiplication by $a_1$. The map is one-to-one and since $S$ is finite, it is injective as well. I understand this, but then the poster goes on to say that this somehow shows that we have $a_1s$ = $a_1$ for some $s$ $∈$ $S$. I have no idea why this would be the case either. If I can figure this out I think I can do the rest of the proof.

Thanks in advance.

Answer 1

For the neutral, Since $G$ is finite, the order of every of its element is finite if $a\in S$, $a^n=e$ for an integer $n$. For the inverse, $aa^{n-1}=e$ so $a^{n-1}$ is the inverse of $a$.

Answer 2

If $a \in S$, and $S$ is closed under products, then $a,a ^2, a^3, a^4, \ldots$ are all in $S$. The group $G$ is finite and hence so is $S$, and so for some $n \neq m$ we have a repetition: $a^n = a^m$. Assume $n > m$ WLOG. Then in $G$ it holds that in fact $a^{n-m}=1$, the neutral element, and as it is a power of $a$, $1 \in S$. Also $a^{n-m-1}=a^{-1}$ is then clear, so $a^{-1} \in S$ too. As $a$ was arbitrary, $S$ is a subgroup.

Answer 3

I would suggest a different approach, OP.

Take any element $a \in S$. Assume WLOG that $a \not = 1$. Then we want to show that both 1 and $a^{-1}$ is also in $S$ if $S$ is closed under composition, and that suffices to show that $S$ is a group. However, for every positive integer $n$, the element $a^n$ is also in $S$. But, as $G$ is finite, so is $S$, so it follows that there exists a $j_1$ and a $j_2 > j_1+1$ such that $a^{j_1} = a^{j_2} =a^{j_1} a^{j_2-j_1} \in S$ [why]. But the equation $a^{j_2}=a^{j_1}$ implies the equation $a^{j_2-j_1} = 1$ [why?] and $a^{j_2-j_1-1} = a^{-1}$. But $a^{j_2-j_1-1}$ and $a^{j_2-j_1-1}$ are both also in $S$ [why]. So 1 is in $S$ and every element $a \in S$ has its inverse also in $S$.

Answer 4

Suppose $A_1: S \to S$ is the map $x \mapsto a_1x$. Then as you noted, the map is one to one, and since $S$ is finite, must also be onto. Hence $a_1$ is in the image of $A_1$ and so there is some $s \in S$ so that $A_1(s)=a_1$. But this is exactly the statement that $a_1s=a_1$ for some $s$.

Answer 5

Consider the positive powers $s^k$ of an element

$s \in S; \tag 1$

since $S$ is closed under the group operation, we have

$s^k \in S, \forall k \in \Bbb N \tag 2$

as well; now since

$\vert G \vert < \infty, \tag 3$

we have

$s^{\vert G \vert} = e, \tag 4$

which in light of (2) is sufficient to show that

$e \in S; \tag 5$

now by virtue of (4), there is a least natural number, denoted by $\vert s \vert \in \Bbb N$, such that

$s^{\vert s \vert} = e; \tag 6$

if now

$\vert s \vert = 1, \tag 7$

then

$s^{-1} = s = e \in S, \tag 8$

whereas if

$\vert s \vert > 1, \tag 9$

then

$ss^{\vert s \vert - 1} = s^{\vert s \vert} = e, \tag{10}$

which shows that

$s^{-1} = s^{\vert s \vert - 1} \in S; \tag{10}$

in either case we have

$s \in S \Longrightarrow s^{-1} \in S. \tag{11}$

We have now shown that $S$, being closed under the group operation by hypothesis, also contains $e$ and $s^{-1}$ for any $s \in S$; thus $S$ satisfies the group axioms and hence is a subgroup of $G$.

Answer 6

I recently posted an answer to a similar question here: https://math.stackexchange.com/a/3116554/625467

I will copy and paste the body here for convenience. In my answer, the subgroup was called H, not S. I will leave out the part where I prove inverses since you mention that is the tiny step you need help getting past and after that you feel you can do it on your own:

"We will show that H is a group. Since H is non-empty, fix $a \in H$. Define $\varphi_a : H \rightarrow H$ by $\varphi_a(h) = ah$ for all $h \in H$. This really does map from H to H because H is closed under the group operation. We will show that $\varphi_a$ is bijective.

Let $h_1, h_2$ be elements of $H$. Suppose $\varphi_a(h_1)=\varphi_a(h_2)$. $$\varphi_a(h_1)=\varphi_a(h_2)$$ $$ah_1=ah_2$$ $$h_1=h_2$$

So $\varphi_a$ is an injective function from one finite set to itself, and is therefore bijective. We can construct this function for any element $a \in H$. This will allow us to show the three group axioms.

1. Associativity: Trivial since $H \subseteq G$

2. Identity: Let $a \in H$ Define $\varphi_a$ as above. From above, this function is bijective, so in particular it is surjective. Surjective means that for all $y \in H$ (H being used here as the co-domain), there is is $x \in H$ (here being used as the domain) such that $\varphi_a(x)=y$. So we apply this property using the fact that we know that $a \in H$. Since $a \in H$, there is some $e \in H$ such that: $$\varphi_a(e)=a$$ $$ae=a$$ And so $e$ must be the identity of G,and must be in H. "

I leave the part about proving that the subset is closed with respect to inverses to you.