# Quick way to determine characteristic subgroups

Reminder: A characteristic subgroup of a group $$G$$ is a subgroup which is stable under all elements of $$\mathrm{Aut}(G)$$. This is a stronger property than being normal.

A while ago there was a question here, which unfortunately got deleted, about a group having two different but isomorphic characteristic subgroups. (It was not this question I found via search, although it basically asks the same thing.)

Of the examples proposed in the comments to that deleted question, the following got stuck in my head:

$$G := Q_8 \times \mathbb Z/8$$

where $$Q_8$$ is the quaternion group with eight elements. (I'll write elements of $$G$$ like $$(-i, 3)$$, i.e. multiplicatively on the left and additively on the right, sorry if that is non-standard.)

The commenting user there claimed $$G$$ has two (and I think they meant: at least two) distinct characteristic subgroups of order $$2$$. Thinking about this, I saw three elements of order $$2$$ in $$G$$, namely

$$(1,4), (-1,0), (-1,4)$$

That the first one is fixed by all automorphisms follows easily from the fact that it is of the form $$g^4$$ for all elements $$g\in G$$ with $$\mathrm{ord}(g)=8$$.

But I needed more work to convince myself that the second and third are also fixed by all automorphisms. Sure it's true for both as soon as it is for one of them; and if an automorphism did not fix them, it would map them to each other; but still to actually show this cannot happen, the best I came up with was writing out what such an automorphism mapping $$(-1,0) \mapsto (-1,4)$$ could possibly do on all $$(\pm i, j, k,0)$$ and eventually getting a contradiction from that. That was not satisfying and to be completely honest, I'm not even sure my proof is correct.

So I wondered if there is a shorter / smarter way to see this, which also applies to more general situations.

How would you determine all characteristic subgroups of the above $$G$$?

As a proposal, is there a way to find the characteristic subgroups of $$G_1 \times G_2$$ if one knows both the characteristic subgroups of $$G_1, G_2$$ and the homomorphisms in either direction $$\mathrm{Hom}(G_i, G_j)$$ (that these hom-sets are important I infer from When is a centerless group characteristic in direct product with $\mathbb{Z}^n$? and common mathematical sense; and in our example, they are very easy to see).

One thing I noticed in this example is that the centre of $$G$$ (which of course is characteristic) is $$\pm1 \times \mathbb Z/8$$, which contains all elements of order $$2$$; but this centre does have an automorphism mapping $$(-1,0) \mapsto (-1,4)$$, which is why I gave up this route and started trying with elements involving $$\pm i,j,k$$.

Would there, in this example or a general setting, have been any way to use information about the centre to conclude?

I'll start by answering your original question: why are the three involutions in $$Q_8\times C_8$$ 'different'? The answer to this is easy: one involution is the power of an element of order $$8$$, and another involution is the one in the derived subgroup. The third is neither of these.

This now means that it's a little easier to determine all characteristic subgroups of $$G$$. We know that any automorphism must fix all involutions, and thus we merely need to consider overgroups of them.

There are obvious characteristic subgroups: each involution, and their direct product. The set of all elements of order dividing $$4$$, the cyclic subgroup of order $$8$$, and so on. Since the automorphisms of the $$Q_8$$ factor extend, you can use them to eliminate some subgroups.

But in general I believe it is quite difficult to determine the full automorphism group. In this case, the outer automorphism group has order $$192=2^4\cdot 3$$, although I checked that in GAP. We can see a group of order $$6$$ acting on $$Q_8$$, and a group of order $$4$$ on the $$C_8$$ factor. So there's an extra $$2$$ that we have missed.

That one isn't so easy to see. It must preserve the $$C_8$$ factor, but cannot preserve the $$Q_8$$ factor as if it preserved both, it could be chosen (via multiplying by the other automorphisms) to centralize both, a contradiction. Thus it maps the $$Q_8$$ to a diagonal subgroup.

For your question about determining the automorphisms of $$G_1\times G_2$$ from having all of the homomorphisms among and between the factors, I don't think that's clear. It doesn't even seem to work in the case of a direct product of isomorphic simple groups. If $$G_1=G_2=C_p$$ then you make up a whole $$\mathrm{GL}_2(p)$$ of outer automorhisms. But if $$G_1=G_2=A_5$$, for example, you only obtain a $$D_8$$ of outer maps.

• Ah of course, the derived subgroup $G' = (\pm1,0)$. Should have looked at that and not just the centre. Aug 6, 2020 at 17:54
• @TorstenSchoeneberg There is an automorphism that almost does what you might think, swapping the two involutions, as I note at the end of my (edited) post. It's just that it maps elements of order $4$ in $Q_8$ to their product with the involution in $C_8$, not the element of order $4$. The same happens with $Q_8\times C_4$. Aug 6, 2020 at 18:11