1. ## Subgroup

(a) Let $(G,*)$ be a finite group. Prove that if $H$ is a nonempty subset of G that is closed under $*$, then $H$ is a subgroup of $G$.

(b) Show that the result in (a) is false if $(G,*)$ is an infinite group.

My attempt for for part (a):

Given:
$H\not = \varnothing$,
$H \subset G$,
$H$ is closed under $*$,
and $G$ is a finite group.

Proof:

If we can show $H$ satisfies $4$ properties; namely, that it is closed under $*$; it is associative; it has an identity, and each of its element has an inverse, then H is a subgroup of $G$.

Since $H$ is closed under $*$, then if $a, b \in H$, then $a*b \in H$. Let $a,b,c \in H$. Since $H \subset G$, $a,b,c \in G$, $(a*b)*c = a*(b*c)$, so $G$ is associative, and $H$ as well.

For all $b \in H$, there is an inverse $b^{-1} \in G$.

Here run into difficulties: Without knowing how to show $b^{-1}\in H$ , I could go nowhere.

For part (b), I find it hard to believe that an infinite group cannot have a subgroup.

2. Part b doesn't mean that infinite groups don't have any subgroups -- it means that the condition, which you have proved sufficient for finite groups, does not work for infinite groups ie. if $(G, \cdot)$ is an infinite group and $H \subseteq G$ is closed under $\cdot$ then $(H, \cdot)$ is not neccesarily a subgroup of $G$.

Now, regarding part a:
Let $h \in H$. $H$ is closed under * and so $h^n \in H \ \forall n \in \mathbb{N}$. $H$ is finite , thus $o(h)$ is finite (this is also why this condition is not sufficient for infinite groups) and both conclusions easily follow (inverse and identity).

3. Dr. Math says, there is a simple way to proof that a subset $H$ of $G$ is a subgroup by "One Step Subgroup Test;" i.e, if $a*b \in H$, then $a*b^{-1}\in H$. This sure makes it a lot easier.

If that's the case, we can say since for $a,b \in H$,
$a*b\in H \Rightarrow a*b^{-1}\in H$ .
Since for all $b\in H$, there is $b^{-1}\in H$, and since $a*b\in H$, it follows that $b*b^{-1}=e \in H$, and so
$H$ is a subgroup of $G$.

Since $H\not = \varnothing$, there exist $x\in H$ for which $x\cdot e = e\cdot x = x$ for all $x\in H$

What do you think?

4. I think you mean "if $a,b \in H$ then $a \cdot b^{-1} \in H$". This is equivalent to what you're trying to prove, since if you prove this then:

1) You prove that $H$ is closed under $*$
2) You prove that $e \in H$ (why?)
3) You prove that $h \in H \Rightarrow h^{-1} \in H$ (why?)

Also, your last statement is illogical - first you're saying that there exists such an $x \in H$ (ie. you take a specific x) , but then you say that the statement is true for all $x \in H$!

5. Originally Posted by Defunkt
I think you mean "if $a,b \in H$ then $a \cdot b^{-1} \in H$". This is equivalent to what you're trying to prove, since if you prove this then:

1) You prove that $H$ is closed under $*$
2) You prove that $e \in H$ (why?)
3) You prove that $h \in H \Rightarrow h^{-1} \in H$ (why?)
I did this because nothing is said about $H$ being a group. It only said $H \not= \varnothing$ and $H \subset G$. Being a subset of $G$ does not necessarily make it a group.

In this case, I don't need to prove that $H$ is closed because it's given.

Also, your last statement is illogical - first you're saying that there exists such an $x \in H$ (ie. you take a specific x) , but then you say that the statement is true for all $x \in H$!
Yes, $\exist x$ and $\forall x$ are contradiction.

I wrote this step because I felt that $H$ has an identity only if $x\cdot e = e \cdot x = x, \forall x \in H$.

By the way, in another book, the Subgroup Test says:

$H\not=\varnothing, H\subset G$ where $G$ is a group, then $H$ is a subgroup of $G$ if and only if $ab\in H$ and $a^{-1} \in H$.

6. Originally Posted by Defunkt
Part b doesn't mean that infinite groups don't have any subgroups -- it means that the condition, which you have proved sufficient for finite groups, does not work for infinite groups ie. if $(G, \cdot)$ is an infinite group and $H \subseteq G$ is closed under $\cdot$ then $(H, \cdot)$ is not neccesarily a subgroup of $G$.

Now, regarding part a:
Let $h \in H$. $H$ is closed under * and so $h^n \in H \ \forall n \in \mathbb{N}$. $H$ is finite , thus $o(h)$ is finite (this is also why this condition is not sufficient for infinite groups) and both conclusions easily follow (inverse and identity).
I can see there be inverses and identity if $h^n \in H \forall n \in \mathbb{Z}$.
Don't we need the negative numbers for inverses so that $h^{-z}\cdot h^{+z}=e$?
Don't we need ${0}\in \mathbb{Z}$ so that $h^0=1$ for identity?

Finally, what's $o(h)$?

7. Originally Posted by novice
I did this because nothing is said about $H$ being a group. It only said $H \not= \varnothing$ and $H \subset G$. Being a subset of $G$ does not necessarily make it a group.
You did what? Sorry, I didn't follow.

By the way, in another book, the Subgroup Test says:

$H\not=\varnothing, H\subset G$ where $G$ is a group, then $H$ is a subgroup of $G$ if and only if $ab\in H$ and $a^{-1} \in H$.
That is virtually the same test. The only difference is that here, you need to explicitly check for the inverse.

8. Originally Posted by novice
I can see there be inverses and identity if $h^n \in H \forall n \in \mathbb{Z}$.
Don't we need the negative numbers for inverses so that $h^{-z}\cdot h^{+z}=e$?
Don't we need ${0}\in \mathbb{Z}$ so that $h^0=1$ for identity?

Finally, what's $o(h)$?
$o(h)$ is the order of $h \in H$, ie. the smallest integer $n \in \mathbb{N}$ such that $h^n = e$.

Since the order of $h$ is finite and $h^n \in H \ \forall n \in \mathbb{N}$, you can write $h^{o(h)} = h^{o(h) - 1 + 1} = h^{o(h)-1} \cdot h = e$ and so $h^{o(h)-1} = h^{-1}$ and $h^{o(h)-1} \in H$.

After you have that $h^{-1} \in H \ \forall h \in H$ and that $H$ is closed under multiplication, identity is easy..

9. Originally Posted by Defunkt
You did what? Sorry, I didn't follow.
Sorry, I meant I did those two things, showing the inverse and identity, because we didn't know whether $H$ is a group or not until it is shown that it's associative, has identity, and inverses for all elements.

I understand the example you gave has all those things; except I have questions about whether or not $h^n \in H, \forall n \in \mathbb{Z}$.

I ask because I can't see $h^n \inH$, $\forall n \in \mathbb{N}$ would have an inverse or identity.

Don't we need $\mathbb{Z}^{-}$ for the inverses and $h^0$ for identity?

10. Originally Posted by Defunkt
$o(h)$ is the order of $h \in H$, ie. the smallest integer $n \in \mathbb{N}$ such that $h^n = e$.

Since the order of $h$ is finite and $h^n \in H \ \forall n \in \mathbb{N}$, you can write $h^{o(h)} = h^{o(h) - 1 + 1} = h^{o(h)-1} \cdot h = e$ and so $h^{o(h)-1} = h^{-1}$ and $h^{o(h)-1} \in H$.

After you have that $h^{-1} \in H \ \forall h \in H$ and that $H$ is closed under multiplication, identity is easy..
I see it now. You are talking about cyclic group, which I have not learned yet, but luckily, I have read it on Wikipedia. Truly understood your example now. Thanks for being so patient.

11. Originally Posted by novice
I see it now. You are talking about cyclic group, which I have not learned yet, but luckily, I have read it on Wikipedia. Truly understood your example now. Thanks for being so patient.
You're very welcome, but I wasn't talking about cyclic groups at all!

12. Originally Posted by novice
I see it now. You are talking about cyclic group, which I have not learned yet, but luckily, I have read it on Wikipedia. Truly understood your example now. Thanks for being so patient.
Ah, it seems you got a bit confused. Since H is closed to *, we get that $(\forall h \in H)(\forall n \in \mathbb{N})h^n \in H$ (try to prove this). This, of course, does not mean that H is cyclic.

We want to show that H is a subgroup of G. In order to show this, we want to show that H has the identity, is closed under * (this we are already given) and that each element in H has an inverse.

To show each element has an inverse - since H is a finite group, every element in it has finite order. Let $h \in H$ and let $o(h)$ be the order of $h$. Then $h^{o(h)} = e$ but also $h^{o(h)} = h^{o(h)-1+1} = h^{o(h)-1}\cdot h = h \cdot h^{o(h)-1}$

That is, for any $h \in H, ~ e = h\cdot h^{o(h)-1}=h^{o(h)-1}\cdot h$ and therefore the inverse of $h$ is $h^{o(h)-1}$. Now it is just left to show that the identity element is in H (I will leave that for you). This concludes (a).

For (b), can you think about some subset of $(\mathbb{Z},+)$ that would be closed under addition but not have the identity element?