# Question about Isomorphism of Aut(G)

In material supplied by my instructor there is a question which asks to pick the incorrect statement among the following:

1. If $$\text{Aut}(G_1)\cong \text{Aut}(G_2)$$ and $$G_1$$ is infinite group then $$G_2$$ is also infinite

2. If $$\text{Aut}(G_1) \cong \text{Aut}(G_2)$$ and $$G_1$$ is finite group then $$G_2$$ is also finite

3. If $$G_1$$ not isomorphic to $$G_2$$ then Aut($$G_1$$) not isomorphic to Aut($$G_2$$).

$$G_1$$ and $$G_2$$ are two groups, Aut(G) is group of their automorphisms and "$$\cong$$" means isomorphism.

I know all three above are incorrect as Aut($$\Bbb Z_3$$)=$$U_3$$ (that is, $$\Bbb Z_2$$) and I also found this statement on groupprops: "Of the three endomorphisms, two are automorphisms: the identity map and the square map. These form a cyclic group of order two: the square map, applied twice, gives the identity map" and Aut($$\Bbb Z$$) is isomorphic to $$\Bbb Z_2$$ so

1. $$G_1=\Bbb Z$$ and $$G_2=\Bbb Z_3$$
2. $$G_2=\Bbb Z$$ and $$G_1=\Bbb Z_3$$
3. We can easily see $$\Bbb Z$$ and $$\Bbb Z_3$$ are not isomorphic but Aut($$\Bbb Z$$) and Aut($$\Bbb Z_3$$) are.

But the given answer is that (1.) is only incorrect statement.

Is there any problem in my reasoning?

• Your reasoning is solid. – Cheerful Parsnip Dec 22 '18 at 7:26