# Thread: automorphism proof

1. ## automorphism proof

If G is a group for which Aut(G)={1}, prove that $|G| \leq 2$

2. Caveat: I am new to abstract algebra.

I can't think of a slick proof. Essentially I show this case by case.

Case 1: $|G|=1$
Because the order of G is 1, then the only element of G is the identity element, G={1}. Further more the only automorphic map $\phi:G\rightarrow G$ is then the identity map.

Case 2: $|G|=2$
Because the order of G is 2, then the only two elements of G are the identity element and some other element, G={1,a}. Because a homomorphism must contain all the inverses for its elements, $a$ must be its own inverse, $a^2 = 1$. Furthermore, because a homomorphism maps identities to identities and inverses to inverses the only possible homomorphism (and therefore automorphism) is again the identity map.

Case 3: $|G|>2$
We need to think up an automorphism that is not the identity map. We need to show that this map is an automorphism. That is as far as I got.

Sorry,

hope this helps some

3. My thinking was for non-abelian groups, conjugation by an element not in the centre. For abelian groups for which not all elements are their inverses, $\theta (x) = x^{-1}$ and for groups where all elements are self inverse, permuting elements should work.

4. Originally Posted by n0083
Caveat: I am new to abstract algebra.

I can't think of a slick proof. Essentially I show this case by case.

Case 1: $|G|=1$
Because the order of G is 1, then the only element of G is the identity element, G={1}. Further more the only automorphic map $\phi:G\rightarrow G$ is then the identity map.

Case 2: $|G|=2$
Because the order of G is 2, then the only two elements of G are the identity element and some other element, G={1,a}. Because a homomorphism must contain all the inverses for its elements, $a$ must be its own inverse, $a^2 = 1$. Furthermore, because a homomorphism maps identities to identities and inverses to inverses the only possible homomorphism (and therefore automorphism) is again the identity map.

Case 3: $|G|>2$
We need to think up an automorphism that is not the identity map. We need to show that this map is an automorphism. That is as far as I got.

Sorry,

hope this helps some
Thanks. I'm new to it also. I had a proof by using cases but i'm stuck on the last case also.

5. so, we may assume that G is an abelian group and every element $\neq 1$ of G has order 2. looking at G as an additive group, now we can consider G as a vector space over $\mathbb{F}_2.$

if $|G| > 2,$ then $\dim_{\mathbb{F}_2} G > 1.$ let $H=\{g_{\alpha}: \ \alpha \in I \}$ be a basis for G. since $|H| \geq 2,$ we may choose $g_{\alpha} \neq g_{\beta} \in H.$ now define $\varphi:G \longrightarrow G$ by $\varphi(g_{\alpha})=g_{\beta}, \ \varphi(g_{\beta})=g_{\alpha}$ and

$\varphi(g_{\gamma})=g_{\gamma},$ for any $\gamma \in I - \{\alpha, \beta \}.$ then $\varphi$ is a non-trivial automorphism of G. Q.E.D.