# Thread: Groups - U10 x U10

1. ## Groups - U10 x U10

Let $G=U_{10} \times U_{10}$ , while $U_{10}$ is the group of all invert elements in $Z_{10}$ , and $U_{10}=\{1,3,7,9\}$ .

1. It is true that Inn(G)={Id} only?

2. How can I know if $Aut(G) \simeq Aut(U_{10})\times Aut(U_{10})$?

Thanks

Let $G=U_10 \times U_10$ , while $U_10$ is the group of all invert elements in $Z_10$ - $U_10=\{1,3,7,9\}$ .

1. It is true that Inn(G)={Id} only?

2. How can I know if $Aut(G) \simeq Aut(U_10)\times Aut(U_10)$?

Thanks
(1) Yes. Your group G is abelian, so what does conjugation do in G? (Remembering that Inn(G) is the set of automorphisms corresponding to conjugation).

For (2) you should realise that an automorphism is defined by where you send your generators to. As a hint, note that switching generators is an automorphism ( $\phi3, 1)\mapsto (1, 3), (1, 3) \mapsto (3, 1)" alt="\phi3, 1)\mapsto (1, 3), (1, 3) \mapsto (3, 1)" /> is an automorphism).

3. Originally Posted by Swlabr
(1) Yes. Your group G is abelian, so what does conjugation do in G? (Remembering that Inn(G) is the set of automorphisms corresponding to conjugation).

For (2) you should realise that an automorphism is defined by where you send your generators to. As a hint, note that switching generators is an automorphism ( $\phi3, 1)\mapsto (1, 3), (1, 3) \mapsto (3, 1)" alt="\phi3, 1)\mapsto (1, 3), (1, 3) \mapsto (3, 1)" /> is an automorphism).
I understand (1). In (2), I can see that the generators of the original group are: 3 and 7. Then, the generators of the second group have to be (3,7), (7,3) :O How do I get on with it?

I understand (1). In (2), I can see that the generators of the original group are: 3 and 7. Then, the generators of the second group have to be (3,7), (7,3) :O How do I get on with it?
The group which you are saying is the Automorphism group corresponds precisely to automorphisms of the form $\phi: (a, b)\mapsto (a\phi_1, b\phi_2)$ where $\phi_i: U_{10}\rightarrow U_{10}$ (can you see why this is?). These automorphisms do not allow for interplay between the a and the b. Specifically, the automorphism I posted above is not of this form.

5. I just don't understand it

First, you said that switching generators is an automorphism, and you showed as an example an automorphism that put 1 in 3, and 3 in 1 (the rest stay the same), but 1 is not a generator of U10 (when U10's action is multiplying) - or did you mean just to switching the places of one or more generator with other elements in U10?

Notice that $G = U_{10} \times U_{10}$ is two-generated. It is generated by the elements $a=(1, 3)$ and $b=(3, 1)$. Define the automorphism $\phi: G \rightarrow G$ by,

$\phi: a\mapsto b$ and $\phi: b \mapsto a$.

7. Hmm.. Let's look at it in a different way.

If I wish to find out how many Automorphisms exist on U10, then after knowing that U10 is cyclic -and- finite, I can know that it has Phi(|O(x)|), while Phi is Euler's function, and O(x) is the group of generators (elements that have the same order as the group itself).

Then, since U10 has two generators (3 and 7), we get that it has 2 automorphisms.

Why does U10 have only 2 generators? What about (1,7)x(7,1)? or maybe even (1,7)x(3,7), and so on? (4 combinations) I got a little confused now

8. The group $U_{10}$ has two choices of generators but it only need one to generate it, while the group $U_{10} \times U_{10}$ need two generators to generate it (for example, $(1, 3)$ and $(3, 1)$).

9. ==Sry, double posting==

10. Originally Posted by Swlabr
The group $U_{10}$ has two choices of generators but it only need one to generate it, while the group $U_{10} \times U_{10}$ need two generators to generate it (for example, $(1, 3)$ and $(3, 1)$).
You mean, I was right in everything else I said ?!

Anyway, I know that U10 has 1 automorphism (which means that Aut(U10)xAut(U10) has ONE element). U10xU10 has more automorphisms (since (1,3),(3,1) are generators, and (1,7),(7,1) are generators, as well as (1,7),(3,1) and (3,1),(1,7)) - I think it has Phi(4) automorphisms, which is 2.

In total, we get that one group is certainly bigger than the other, so it'll never be isomorphic.

Is this right?

You mean, I was right in everything else I said ?!

Anyway, I know that U10 has 1 automorphism (which means that Aut(U10)xAut(U10) has ONE element). U10xU10 has more automorphisms (since (1,3),(3,1) are generators, and (1,7),(7,1) are generators, as well as (1,7),(3,1) and (3,1),(1,7)) - I think it has Phi(4) automorphisms, which is 2.

In total, we get that one group is certainly bigger than the other, so it'll never be isomorphic.

Is this right?
You just proved in your previous post that $U_{10}$ has two automorphisms! Not one!!!

So, $Aut(U_10) \times Aut(U_{10})$ has four elements. These are the following four automorphisms. Note that I give them in terms of where the generators (1, 3) and (3, 1) are sent to,

$\phi_0: (1, 3) \rightarrow (1, 3), (3, 1) \mapsto (3, 1)$
$\phi_1: (1, 3) \rightarrow (1, 7), (3, 1) \mapsto (7, 1)$
$\phi_2: (1, 3) \rightarrow (1, 3), (3, 1) \mapsto (7, 1)$
$\phi_3: (1, 3) \rightarrow (1, 7), (3, 1) \mapsto (3, 1)$.

Can you come up with any more automorphisms of this group?

12. Oh, cool, I just got a little 'sharper' lately in this whole 'count the automorphisms' thing, after taking a group, such as U7, and trying to find the automorphisms and realize why it's so. I'm still looking for a 'shortcut', using Euler's function, in any general case.

Thanks

Oh, cool, I just got a little 'sharper' lately in this whole 'count the automorphisms' thing, after taking a group, such as U7, and trying to find the automorphisms and realize why it's so. I'm still looking for a 'shortcut', using Euler's function, in any general case.

Thanks
(Note that I use $C_{n}$ for the cyclic group of order $n$).

Such a short-cut doesn't exist, because the order of an automorphism group is not dependent on the order of the group.

For example, $G = U_{10} \times U_{10} \cong C_3 \times C_3$, $|G| = 9$, and $|Aut(G)| = 8$ (I believe) but $8=|Aut(G_1)|>|Aut(C_9)| = 6$, the automorphism group of the cyclic group or order 9. Thus, we have two groups of the same order but with different automorphism groups of different orders.

However, I believe the solution for abelian groups isn't too hard. It should, I feel, look something like,

Let $G=C_{x_1}^{y_1}\times C_{x_2}^{y_2}\times \ldots times C_{x_n}^{y_n}$ be an abelian group. There is a theorem which says that every abelian group is of this form. Then $|Aut(G)| = (\phi({x_1})^{y_1})^{y_1!}(\phi({x_2})^{y_2})^{y_2 !}\ldots(\phi({x_n})^{y_n})^{y_n!}$ where the first power deals with the automorphism which keep the generators in the same position in the direct product, while the $y_i!$ moves them around. $\phi$ is Euler's totient function.

14. Originally Posted by Swlabr
(Note that I use $C_{n}$ for the cyclic group of order $n$).

Such a short-cut doesn't exist, because the order of an automorphism group is not dependent on the order of the group.

For example, $G = U_{10} \times U_{10} \cong C_3 \times C_3$, $|G| = 9$, and $|Aut(G)| = 8$
Didn't we just say before that U10xU10 has two automorphisms?

Didn't we just say before that U10xU10 has two automorphisms?
No. U_{10} has STRICTLY more than 4 automorphisms. I wrote down 4 of them in post no. 11 or so. I gave them in terms of the generators, as an automorphism is defined precisely by where the generators are sent to (can you see why this is?).

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