# Automorphism

• Aug 11th 2010, 10:25 PM
demode
Automorphism
Let $\displaystyle G = Aut(\mathbb{Z}_{315})$ and let $\displaystyle n$ be the order of the largest cyclic subgroup of G. What is $\displaystyle n$? And what would be the number of distinct cyclic subgroups of order $\displaystyle n$ in $\displaystyle G$?
• Aug 11th 2010, 11:00 PM
Swlabr
You should start by thinking about what this automorphism group looks like. Essentially, this comes down to where can the generator, 1, be sent to by an automorphism?'. (e.g. $\displaystyle \phi: 1 \rightarrow 2$ is an automorphism. Can you think why this is?)

So, where can the generator be sent to?
• Aug 14th 2010, 04:21 AM
demode
Quote:

Originally Posted by Swlabr
You should start by thinking about what this automorphism group looks like. Essentially, this comes down to where can the generator, 1, be sent to by an automorphism?'. (e.g. $\displaystyle \phi: 1 \rightarrow 2$ is an automorphism. Can you think why this is?)

So, where can the generator be sent to?

The generator is sent to $\displaystyle U(315)$, as $\displaystyle Aut(\mathbb{Z}_{315})$ is isomorphic to $\displaystyle U(315)$?

I think the order of the largest cyclic subgroup of G is

For an element $\displaystyle \alpha$ of $\displaystyle Aut(\mathbb{Z}_{315})$ we know $\displaystyle \alpha(1)$, we'll know $\displaystyle \alpha(k)$ fora any $\displaystyle k$, because

$\displaystyle \alpha(k) = \alpha \underbrace{(1+1+...+1)}_\textrm{k terms} = \underbrace{\alpha(1)+...+\alpha(1)}_\textrm{k terms}=k\alpha(1)$

I think the property that isomorphisms preserve orders would tell us that $\displaystyle |\alpha(1)|=315$, then I think members of $\displaystyle U(315)$ would be the candidates for $\displaystyle \alpha(1)$.

The largest member of $\displaystyle U(315)$ is 314. But I don't know if this helps...
• Aug 14th 2010, 11:39 PM
demode
Oh, I meant since $\displaystyle \phi$ is an automorphism we have $\displaystyle \phi(n)=dn$ for some $\displaystyle d \in U(315)$ and any $\displaystyle n \in Z_{315}$.

But how does this help me to prove the question?
• Aug 16th 2010, 12:47 AM
Swlabr
Yes, you need to find out where you can send the generator too. Your choices are precisely those numbers which generate $\displaystyle \mathbb{Z}_{315}$. So, which elements of your group generate it? How many of them are there?

This will give you the order of your Automorphism group (why?).
• Aug 16th 2010, 04:59 AM
demode
Thanks for the reply. Since $\displaystyle |a^k| = \frac{315}{gcd(315,k)}$, $\displaystyle G = \left\langle a \right\rangle \iff k \in U(315)$, where

$\displaystyle U(315) = \{1,2,4,6,...,314 \}$

Therefore the generators of $\displaystyle \mathbb{Z}_{315}$ are the elements

$\displaystyle \{ a,a^2,a^4,a^6...,a^{314} \}$

Is this right? If so how do I know how many elements there are in $\displaystyle U(315)$? (there are too many to count)
• Aug 21st 2010, 01:25 PM
demode
Okay I know that if G is a cyclic group of order n, then $\displaystyle Aut (G) \cong U(n)$.

In our case if we have $\displaystyle \mathbb{Z}_{315}= \left\langle a \right\rangle$

$\displaystyle Aut(\mathbb{Z}_{315}) = \{ a_i : \mathbb{Z}_{315} \to \mathbb{Z}_{315}, a \mapsto ia, i \in U(315) \} \cong U(315)$

$\displaystyle 315 = 5.7.3^2$

$\displaystyle U(315) \cong \mathbb{Z}_5 \oplus \mathbb{Z}_7 \oplus \mathbb{Z}_9$

Does this mean the order of the largest cyclic subgroup of G is 63? Because $\displaystyle 5 \times7 < 5 \times 9 < 7 \times 9 = 63$.

Am I right?
• Aug 22nd 2010, 11:39 PM
Swlabr
Quote:

Originally Posted by demode
Okay I know that if G is a cyclic group of order n, then $\displaystyle Aut (G) \cong U(n)$.

In our case if we have $\displaystyle \mathbb{Z}_{315}= \left\langle a \right\rangle$

$\displaystyle Aut(\mathbb{Z}_{315}) = \{ a_i : \mathbb{Z}_{315} \to \mathbb{Z}_{315}, a \mapsto ia, i \in U(315) \} \cong U(315)$

$\displaystyle 315 = 5.7.3^2$

$\displaystyle U(315) \cong \mathbb{Z}_5 \oplus \mathbb{Z}_7 \oplus \mathbb{Z}_9$

Does this mean the order of the largest cyclic subgroup of G is 63? Because $\displaystyle 5 \times7 < 5 \times 9 < 7 \times 9 = 63$.

Am I right?

Largely yes, this is all correct. Apart from your answer. There is a cyclic group of order 5.7.3 there...