U(15) automorphisms

Question

I was wondering how to find automorphism group of $ U(15) $. I've read some similar questions and found this one: Automorphisms and Mappings. It's quite helpful, but I don't understand everything.

Could someone please explain me, why is 'any automorphism $ φ $ determined by where we send the two generators $ (1, 0), (0, 1) $'? Does always exist such a generating set of some group that we can say something like this? In general, does sending generators of a group $ G $ determine an automorphism?

Now consider $ U(15) $. We want to find automorphism group of $ U(15) $. Generating set of $ U(15) $ is $ \{2, 11\} $ as well as $ \{7, 4\} $ (just to name a few).

For $ φ(2) $ there are $ 4 $ options: 2, 7, 8, 13 and for $ φ(11) $ there are $ 3 $ options: $ 4, 11, 14 $. However, the first element is actually the square of any of the order $ 4 $ elements. So $ 4 $ is not possible. This gives at most $ 4 \times 2 $ possible automorphisms and that's fine. Now consider generating set $ \{7, 4\} $. For $ φ(7) $ there are $ 4 $ options: $ 2, 7, 8, 13 $ and for $ φ(4) $ there are $ 3 $ options: $ 4, 11, 14 $. However, only first element is the square of any of the order $ 4 $ elements. So $ 11, 14 $ are not possible. This gives at most $ 4 \times 1 $ possible automorphisms and this is less than expected. So can we say there are 'better' and 'worse' generating sets of $ U(15) $ to find automorphism group of $ U(15) $?

I don't understand what's the difference between above generating sets when we want to find automorphism group of $ U(15) $. I would say it's easier to consider $ \{2, 11\} $, not $ \{7, 4\} $. Am I right?

I'm a totally beginner in abstract algebra and don't know any theorems. I'm asking for a simple explanation without theorems or 'well-known facts'.

Answer 1

Generators are building blocks for a group, in the sense every element of the group can be constructed using the generators (if you have some background in Linear Algebra, then think of what a basis set does for the vector space, every vector can be expressed in terms of basis vectors.)

In order to find an automorphism of the group $G=\Bbb{Z}_4 \times \Bbb{Z}_2$, we look for a generating set. The set $\{(1,0), (0,1)\}$ happens to be a generating set for this group because $$\text{for } (a,b) \in G, \quad (a,b)=a(1,0)+b(0,1).$$ Now to find an automorphism $\varphi:G \rightarrow G$, we want to determine $\varphi(a,b)$ for all $(a,b) \in G$. Instead of going through all the elements we use the homomorphic property to determine $\varphi$ as follows: $$\varphi(a,b)=\varphi(a(1,0)+b(0,1))=a\varphi(1,0)+b\varphi(0,1).$$ This tells us if we know the images of $(1,0)$ and $(0,1)$, then we can get the images of all elements of the group. That is why generators are useful.

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Now for $H=U(15)=\{1,2,4,7,8,11,13,14\}$. Actually $\{7,4\}$ is NOT a generating set for $U(15)$ because $\langle 7 \rangle=\{1,7,4,13\}$ and $\langle 4 \rangle=\{1,4\} \subset \langle 7 \rangle$. That's the reason you are not getting the answer.

Since $\langle 2 \rangle=\{1,2,4,8\}$ and $\langle 11 \rangle=\{1,11\}$, so $\langle 2 \rangle \cap \langle 11 \rangle =\{1\}$. Thus $\{2,11\}$ will form a generating set for $H$.

Suppose you want a generating set of $H$ that contains $7$. Then look for an order $2$ element of $H$that does not appear in the subgroup generated by $7$. For example, if we take $\{7,11\}$, then that will work.