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Question is to prove that :

A finite group whose only automorphism is identity map must have order at most $2$.

What i have tried is :

As any automorphism is trivial, so would be inner automorphism

i.e., each map for fixed $g\in G $ with $\eta : G\rightarrow G$ taking $h$ to $ghg^{-1}$ is trivial.

Thus, $ghg^{-1}=g$ i.e., $gh=hg$ for all $g\in G$ and $h\in G$ which would say that $G$ is abelian.

So, I would have that $G$ is finite abelian group.

Now, As $G$ is abelian, the map $g\rightarrow g^{-1}$ is an automorphism.

But only automorphism is identity map, so we would have :

$g=g^{-1}$ i.e., $g^2=e$ for all $g\in G$

So, I would have that $G$ is group with each element of order $2$.

Combining with previous result I would have :

$G$ is a finite abelian group in which each element is of order $2$

I am not able to conclude anything more than this....

A kind of cheating would give something very close :

As group is finite abelian which has each element with order $2$, It should be :

$\mathbb{Z}_2\times \mathbb{Z}_2$ Or

$\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_2$ Or

$\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_2$ Or something very similar to this.

For first group $\mathbb{Z}_2\times \mathbb{Z}_2$ automorphism group is general linear group of order $2$ with entries from $\mathbb{Z}_2$ which is not trivial. So, this should not be the required group.

This would hold for similar cases

So, I feel that i am on right path but i need some help to make it more clear.

Thank you :)

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  • $\begingroup$ You basically have it! Your observation that Aut (Z/2 \times Z/2) = GL_2 (Z/2) generalizes in a straigtforward case to the situation where there are n factors of Z/2 on the lhs. What must you change on the rhs? $\endgroup$ – hunter Dec 25 '13 at 13:42
  • $\begingroup$ I should have $GL_n (\mathbb{Z}_2) $ which is not identity so....... :) :) $\endgroup$ – user87543 Dec 25 '13 at 14:02
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A (finite) abelian group with $g^2=e$ for all $g\in G$ is in fact a (finite dimensional) $\Bbb Z_2$-vector space, and then its automorphisms correspond to invertible matrices over $\Bbb Z_2$, so...

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  • $\begingroup$ I got your point.. I am just waiting if some one would give a more general description.... Thank you so much for your interest :) $\endgroup$ – user87543 Dec 25 '13 at 14:02
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You are already very close.

Indeed, nonabelian groups have nontrivial inner automorphisms, while abelian ones have the automorphism $a\mapsto a^{-1}$, which is trivial only when all $a$ have order $2$.

So we only need to deal with the direct sums of $\mathbb{Z}_2$. But for such groups we have the automorphism \begin{equation} (x,y)\mapsto (y,x), \end{equation} which is nontrivial unless we have only one copy in our direct sum.

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  • $\begingroup$ This is so helpful.... I had not recognized your last statement before... Thank you so much :) $\endgroup$ – user87543 Dec 25 '13 at 15:41
  • $\begingroup$ Yes, these are cute little facts XD $\endgroup$ – Hui Yu Dec 25 '13 at 15:58
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We show that $Aut(G)$ is nontrivial if $G$ is abelian, $|G|>2$ (may be infinite), and all its non-identity elements are of order 2.

Let $G$ be an abelian group with all non-identity elements of order $2$ and $|G|>2$. It's easy to see that $A = \{1, a_1, a_2, a_1a_2\}$ is a subgroup of $G$, where $a_1$ and $a_2$ are distinct non-identity elements of $G$. Let $B$ be a subgroup of $G$ and also a transversal of the collection of all cosets of $A$. Let $h$ be a nontrivial automorphism on $A$, and we can cronstuct an nontrivial automorphism $f$ on $G$ by defining $f$($b$$a$) = $b$$h$($a$) for all $b \in B$ and $a \in A$. $ So, Aut(G)$ is nontrivial.

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