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

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.

• Set theory fact: if $S$ is a finite set and $f:S \to S$ is a function then $f$ is 1-1 iff $f$ is onto iff $f$ is a bijection. – Henno Brandsma Feb 16 at 23:06
• See Theorem 3.3 of Gallian's "Contemporary Abstract Algebra (Eighth Edition)". – Shaun Feb 16 at 23:10

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.

• I've seen this proof as well but I am also confused about this. I understand that $S$ is closed under the group operation, and that for it to be finite some condition has to be satisfied $(a^m = a^n)$ $m ≠ n$ but I can't picture why this is true. Why does this fact allow for the set to still be closed under group operation and for it to be finite? The answer may be trivial (hence why people don't go very into depth about it) but I am having a hard time understanding it. – ForIgreaterthanJ Feb 18 at 3:16
• @ForIgreaterthanJ because otherwise the function $a \to a^n$ from $\mathbb{N}$ to $S$ would be injective (1 to 1) and by standard set theory this implies that $S$ is infinite. Contradiction, so the function cannot be injective and we have such a repeat value. – Henno Brandsma Feb 18 at 5:00
• That makes perfect sense. Thank you very much for your help. – ForIgreaterthanJ Feb 18 at 15:47

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 $$s$$.

• Wow, it just clicked. Thank you so much! I am still trying to understand the other methods to prove this but you helped a lot with this method. Thanks again! – ForIgreaterthanJ Feb 18 at 3:38
• Great! I'm glad it makes sense. – Alexander Feb 18 at 6:12
• If I may, I have another quick question. I intuitively understand what it means to say that if $A_{1}$ is injective on a finite set then it is also surjective, but I was going back over this proof again so I could post it for proof verification and I ran into another problem. Apart from the intuitive understanding, how do we actually go about proving $A_1$ is surjective? It seems that for $A_1$ to be surjective the pre-image would need to be $s = a^{-1} y$ s.t. $A_1(s) = y$. But am I allowed to use inverses here? $S$ hasn't been shown to be closed under inverses yet. – ForIgreaterthanJ Feb 21 at 19:38
• The point is just that $A_1: S \to S$ is injective and so the image $A_1(S) \subset S$ is a subset of $S$ containing exactly $|S|$-elements. Thus, $A_1(S)=S$. This shows $A_1$ is surjective but note we didn't need to talk about inverses. – Alexander Feb 22 at 3:08
• Perfect. Again, thank you very much. I haven't done very much math since calculus in college so I am a little rusty. But that definitely makes sense. Again, thank you! – ForIgreaterthanJ Feb 22 at 17:35

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 $$a$$.

• You should explain why the order of every element is finite - this is the part the OP doesn't seem to understand.. – Mariah Feb 16 at 23:28

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 $$G$$.

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 $$S$$.

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.