1. ## Automorphism

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 $G$?

2. 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?

3. 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...

4. 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?

5. 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?).

6. 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)

7. 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?

8. 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...