Is this a valid strategy to find group new group automorphisms given you already know some?

Question

It is well known that Aut($G$), the group of all automorphisms of a group $G$, is a group under function composition. Lets say that you know $n$ of the automorphisms (plus the identity isomorphism) on $G$, hence you have $$H=\{id, \phi_1, \phi_2,\dots,\phi_n\}$$ Lets also say we do not know if $H = \text{Aut}(G)$ yet. I thought of the following strategy to come up with new Automorphisms:

Check if $H$ is a group. If not, take some $\phi_i,\phi_j \in H$ and find $\phi_i \cdot \phi_j$. If $\phi_i \cdot \phi_j$ is not in $H$, we have found another automorphism on $G$; add it to $H$. If it is not a new automorphism, pick new $\phi_i, \phi_j$. Repeat this process until $H$ is a group.

My question is, is this even a good strategy? One clear problem is that if we arrive at a subgroup of $\text{Aut}(G)$, we are obviously not going to generate any more automorphisms of $G$. Is there any way to optimize this strategy? Another thought I had was to take $$\phi_i^2, \phi^3,\dots,id$$ to find new automorphisms instead of composing $2$ randomly chosen ones, but I am not sure if this is better or worse.

Answer 1

Stated somewhat differently: suppose you have some elements $h_1, \ldots, h_n$ of a finite group.

1. Start with $S = \{h_1, \ldots, h_n\}$, and mark them all.
2. Repeat while $S$ has some marked elements:
3. ... Let $a$ be the first marked element $a$ of $S$.
4. .. Place into $S$ all elements $ab$ with $b \in S$ that are not already in $S$, and mark them.
5. .. Remove the mark from $a$.

When done, $S$ will be the subgroup of your group generated by $h_1, \ldots, h_n$.

Answer 2

Unfortunately no, your strategy will not in general produce all automorphisms of a group. Indeed, as the other answer mentions, you end up producing the group generated by the automorphisms that you have. Of course, the question does not specify that $G$ is finite, so you also need to include inverses of elements, as otherwise you only need generate a semigroup (i.e., closed under products but not necessarily inverses).

Constructing $\mathrm{Aut}(G)$ for $G$ an arbitrary group is a very difficult thing. Even for $G$ finite one has to know a lot of group theory to set about doing this, and to know that you have all automorphisms. One incredibly bad case is when $G$ is a $p$-group (i.e., $|G|=p^n$ for some $n$), for example.