Automorphism

**demode** · August 11th 2010, 10:25 PM

Let $G = Aut(\mathbb{Z}_{315})$ and let $n$ be the order of the largest cyclic subgroup of G. What is $n$ ? And what would be the number of distinct cyclic subgroups of order $n$ in $G$ ?

**Swlabr** · August 11th 2010, 11:00 PM

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. $\phi: 1 \rightarrow 2$ is an automorphism. Can you think why this is?)

So, where can the generator be sent to?

**demode** · August 14th 2010, 04:21 AM

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. $\phi: 1 \rightarrow 2$ is an automorphism. Can you think why this is?)

So, where can the generator be sent to?

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

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

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

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

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

The largest member of $U(315)$ is 314. But I don't know if this helps...

**demode** · August 14th 2010, 11:39 PM

Oh, I meant since $\phi$ is an automorphism we have $\phi(n)=dn$ for some $d \in U(315)$ and any $n \in Z_{315}$ .

But how does this help me to prove the question?

**Swlabr** · August 16th 2010, 12:47 AM

Yes, you need to find out where you can send the generator too. Your choices are precisely those numbers which generate $\mathbb{Z}_{315}$ . 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?).

**demode** · August 16th 2010, 04:59 AM

Thanks for the reply. Since $|a^k| = \frac{315}{gcd(315,k)}$ , $G = \left\langle a \right\rangle \iff k \in U(315)$ , where

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

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

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

Is this right? If so how do I know how many elements there are in $U(315)$ ? (there are too many to count)

**demode** · August 21st 2010, 01:25 PM

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

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

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

$315 = 5.7.3^2$

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

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

Am I right?

**Swlabr** · August 22nd 2010, 11:39 PM

Originally Posted by demode

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

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

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

$315 = 5.7.3^2$

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

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

Am I right?

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

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