# Math Help - Cyclic Groups

1. ## Cyclic Groups

The question has 3 parts:

My Attempt:

(a) Since the cyclic group generated by a is $\left\langle a \right\rangle = \{ za | z \in \mathbb{Z} \}$, I think it would follow that for distinct $a_1,a_2,...,a_n \in \mathbb{Z}$

$\left\langle a_1,...,a_n \right\rangle = \{ z_1a_1 +...+z_na_n\}$

but I don't know how to actually prove this. I appreciate any hints to get started.

(b) Though I'm not sure if it helps... but I think in general $\left\langle ma \right\rangle \cap \left\langle na \right\rangle = \left\langle ka \right\rangle$ where $k =lcm(m,n) \mod\ \mathbb{Z}$. Is that correct?

(c) Is the gcd(a,b) required to be 1 (so that a and b are relatively prime)? If so I get

$\left\langle a,b \right\rangle = \{ au+bv | u,v \in \mathbb{Z}\}= \left\langle h \right\rangle$

$\implies au+bv=1$

Is "h" equal to 1?

2. Part (a) is one of those annoying questions which is quite genuinely obvious and so annoying to prove. I would try inducting on 'n', the number of generators. If that doesn't work, I wouldn't worry too much about it!

Yes, in (b) it is the least common multiple you are after.

For (c) you are on the right lines, but generalise your result of gcd=1 to arbitrary integers.

3. Originally Posted by Swlabr
Part (a) is one of those annoying questions which is quite genuinely obvious and so annoying to prove. I would try inducting on 'n', the number of generators. If that doesn't work, I wouldn't worry too much about it!
I plan to prove it in 3 steps but I'm having some trouble:

1. $\forall z_1,...,z_n$, we have $z_1a_1+z_2a_2+...+z_na_n \in \left\langle a_1,...,a_n \right\rangle$

Therefore $\imples \{ z_1a_1+z_2a_2+...+z_na_n | z \in \mathbb{Z} \} \subseteq \left\langle a_1,...,a_n \right\rangle$

The problem is that I don't see how to prove that every linear combination $z_1a_1+z_2a_2+...+z_na_n$ must be in $\left\langle a_1,...,a_n \right\rangle$.

2. $\{ z_1a_1+z_2a_2+...+z_na_n | z \in \mathbb{Z} \}$ is a subgroup of $\mathbb{Z}$ containing $\left\langle a_1,...,a_n \right\rangle$.

To prove that it's a subgroup is it possible to use the "one step test"?

3. If I can then show that $\left\langle a_1,...,a_n \right\rangle$ is the smallest subgroup of $\mathbb{Z}$ containing $a_1,...,a_n$ with $\{ z_1a_1+z_2a_2+...+z_na_n | z \in \mathbb{Z} \} \supseteq \left\langle a_1,...,a_n \right\rangle$

I have shown that

$\{ z_1a_1+z_2a_2+...+z_na_n | z \in \mathbb{Z} \} \subseteq \left\langle a_1,...,a_n \right\rangle$
$\{ z_1a_1+z_2a_2+...+z_na_n | z \in \mathbb{Z} \} \supseteq \left\langle a_1,...,a_n \right\rangle$

$\therefore \{ z_1a_1+z_2a_2+...+z_na_n | z \in \mathbb{Z} \} = \left\langle a_1,...,a_n \right\rangle$

Yes, in (b) it is the least common multiple you are after.
Actually I meant:

$\left\langle a \right\rangle \cap \left\langle b \right\rangle = \left\langle k \right\rangle$

where k = lcm(a,b) mod Z = <g>.

So I think here I have to go both ways i.e: show that

$\left\langle a \right\rangle \cap \left\langle b \right\rangle \leq lcm(a,b)$ .....(1)

$\left\langle a \right\rangle \cap \left\langle b \right\rangle \geq lcm(a,b)$ .....(2)

Could you show me how to go about proving one of them? I absolutely don't know where to start...

For (c) you are on the right lines, but generalise your result of gcd=1 to arbitrary integers
How? It follows from part (a) that

$\left\langle a,b \right\rangle = \{ z_aa+z_bb | z_a,z_b \in \mathbb{Z}\}\in gcd (a,b)$

$\exists z_a, z_b \in \mathbb{Z}$ with $z_a a + z_bb=gcd(a,b)$

I think similar to part (b) I have to prove in both directions:

$\left\langle a,b \right\rangle \leq \left\langle gcd(a,b) \right\rangle$
$\left\langle a,b \right\rangle \geq \left\langle gcd(a,b) \right\rangle$

We can let $h=gcd(a,b)$ then we can write a=ch and b=dh where c,d are integers.

If $x \in \left\langle a \right\rangle \implies x \in \left\langle ch \right\rangle$

Also if $x \in \left\langle b \right\rangle \implies x \in \left\langle dh \right\rangle$

Then what can we say next?

P.S. By the " $\leq$" notation I mean 'a subgroup of', not 'less than or equal to'.

4. But if my method is hopeless, then you could show me how to do it with induction.