# If $G$ has no non-trivial automorphism, then $G$ is abelian and $g^2 = e$ for all $g \in G$ . [duplicate]

If $G$ has no non-trivial automorphism, then $G$ is abelian and $g^2 = e$ for all $g \in G$ .

With the assumption, I dont know how to start the proof.

If there is no non-trivial automorphism, then there is only trivial automorpism, the identity morphism. But how can I show $g^2 = e$ for all $g \in G$ with it?

## marked as duplicate by Jyrki Lahtonen abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Nov 17 '17 at 10:17

• Begin with considering inner automorphisms. – Boris Novikov Apr 25 '13 at 15:38
• Boris said it well...With your assumption, G has no non-trivial automorphism...what does that mean? – Eleven-Eleven Apr 25 '13 at 15:45
• The first step is to show that the group is Abelian, using the fact that the inner automorphisms are trivial. Then, you have won something: if (and only if) a group is Abelian, the inversion $x \mapsto x^{-1}$ is an automorphism. Using the hypothesis, you then have $\forall x \in G, x^{-1} = x$, which is equivalent to all elements of the group being involutions. – PseudoNeo Apr 25 '13 at 16:15

For every elements $a\in G$, consider $$\phi_a:G\rightarrow G$$ sending an element $b$ to its conjugate $a^{-1}ba$. You can easily check that all $\phi_a$ are automorphisms of $G$, called inner automorphisms. By assumption, $\phi_a=id$, which means $$\phi_a(b)=b$$ for every $a,b\in G$, so that $ba=ab$.

Alternatively, use $$\frac{G}{Z(G)}\cong Inn(G)$$ where $Z(G)$ is the center of $G$ and $Inn(G)$ the group of inner automorphisms, which is trivial in your hypothesis.

As for $g^2=e$, you can use abelianity we have just proved. Since $(ab)^{-1}=b^{-1}a^{-1}=a^{-1}b^{-1}$, the inversion $g\rightarrow g^{-1}$ is an automorphism, which by assumption is trivial. Multiplying $g=g^{-1}$ by $g$ gives $g^2=e$

Hint: If $Inn(G)$ is the group of inner isomorphisms, which we are taking to be trivial, apply the following theorem:

$G/Z(G)\cong Inn(G)$.

Recall: If $G$ is Abelian, what is the relationship between $G$ and $Z(G)$?

After the observations above, note that in case $G$ is finite (for infinite groups a similar reasoning applies), it must be an elementary abelian 2-group of rank say $n$. Then $Aut(G) \cong GL(n,2)$. Hence if $|Aut(G)| = 1$ then this can only be the case if $G$ is trivial, or $n = 1$ that is $G \cong C_2$, the cyclic group of order 2.