1. ## Ugh! Modules....

I'll write "$\displaystyle Z$" to denote the set of whole numbers

Let $\displaystyle n > 1$ be a whole number and let $\displaystyle a$ be a fixed whole number. Prove that

H = {x ∈ Z, such that ax ≡ 0 (mod n)}

is a subgroup of $\displaystyle Z$ in the sum.

In the sum is the part where I'm confused... I though you could only multiply with modules...?

2. Originally Posted by Tall Jessica
I'll write "$\displaystyle Z$" to denote the set of whole numbers

Let $\displaystyle n > 1$ be a whole number and let $\displaystyle a$ be a fixed whole number. Prove that

H = {x ∈ Z, such that ax ≡ 0 (mod n)}

is a subgroup of $\displaystyle Z$ in the sum.

In the sum is the part where I'm confused... I though you could only multiply with modules...?
Hi Tall Jessica,

Let $\displaystyle x,y \in H$, check if $\displaystyle xy^{-1} \in H$ and $\displaystyle x^{-1} \in H$ is true.
(Since it is under the operation of "+", $\displaystyle y^{-1}$ is the same as $\displaystyle -y$, so $\displaystyle xy^{-1} =x-y$)

3. Originally Posted by chabmgph
Hi Tall Jessica,

Let $\displaystyle x,y \in H$, check if $\displaystyle xy^{-1} \in H$ and $\displaystyle x^{-1} \in H$ is true.
(Since it is under the operation of "+", $\displaystyle y^{-1}$ is the same as $\displaystyle -y$, so $\displaystyle xy^{-1} =x-y$)

Okay, so...

I multiply both sides by $\displaystyle y^{-1}$, which gives me:

(ax)y^{-1} ≡ 0 (mod n)

(ay^{-1})(xy^{-1}) ≡ 0 (mod n)

(a-y)(x-y) ≡ 0 (mod n)

...is this okay? I'm soooo lost... lol!

4. Originally Posted by Tall Jessica
Okay, so...

I multiply both sides by $\displaystyle y^{-1}$, which gives me:

(ax)y^{-1} ≡ 0 (mod n)

(ay^{-1})(xy^{-1}) ≡ 0 (mod n)

(a-y)(x-y) ≡ 0 (mod n)

...is this okay? I'm soooo lost... lol!
Well, you can't assume that $\displaystyle (xy^{-1}) \equiv 0 \mod n$, can you? That is what you are trying to prove. lol. (I know, I have those moments too.)

You may want to try this:
Since $\displaystyle x,y \in H$, we have
$\displaystyle ax \equiv 0 \mod n$
$\displaystyle ay \equiv 0 \mod n$
Then, $\displaystyle ax - ay \equiv 0 \mod n$
...
Does that help?

5. Originally Posted by chabmgph
Well, you can't assume that $\displaystyle xy^{-1} \equiv 0 \mod n$, can you? That is what you are trying to prove. lol. (I know, I have those moments too.)

You may want to try this:
Since $\displaystyle x,y \in H$, we have
$\displaystyle ax \equiv 0 \mod n$
$\displaystyle ay \equiv 0 \mod n$
Then, $\displaystyle ax - ay \equiv 0 \mod n$
...
Does that help?
Oy, vey! It's NOT my day! lol! I HATE modules! I'll be honest, I can hardly work them.

Let's see...

$\displaystyle ax - ay \equiv 0 \mod n$

I can take out a common factor $\displaystyle a$

$\displaystyle a (x - y) \equiv 0 \mod n$

Which would give me

$\displaystyle a xy^{-1} \equiv 0 \mod n$

...yes?

6. Originally Posted by Tall Jessica
Oy, vey! It's NOT my day! lol! I HATE modules! I'll be honest, I can hardly work them.

Let's see...

$\displaystyle ax - ay \equiv 0 \mod n$

I can take out a common factor $\displaystyle a$

$\displaystyle a (x - y) \equiv 0 \mod n$

Which would give me

$\displaystyle a xy^{-1} \equiv 0 \mod n$

...yes?

So that means $\displaystyle xy^{-1} \in H$, Now don't forget to check if $\displaystyle x^{-1} \in H$ is true.

7. I knew it :-)

I'm awsome! lol!!

Thank you :-)

8. Originally Posted by chabmgph

So that means $\displaystyle xy^{-1} \in H$, Now don't forget to check if $\displaystyle x^{-1} \in H$ is true.

Wait. If $\displaystyle x \in H$ is true, doesn't it mean that $\displaystyle x^{-1} \in H$ is also true?

9. Originally Posted by Tall Jessica
Wait. If $\displaystyle x \in H$ is true, doesn't it mean that $\displaystyle x^{-1} \in H$ is also true?
Yup.
Since ax is divisible by n, -ax = a(-x) is also divisible by n.

10. Originally Posted by chabmgph
Yup.
You're just givin' me the runaround!

You meanie!

(Thanx) :-)