# Group, which doesn't have $\mathrm{Aut}(G)$. [duplicate]

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Does there some groups which doesn't contain $(\mathrm{Aut}(G),*)$ (only $\mathrm{id}$), where $*$ is composition. I thought that $\mathbb{R}$ could be such group, but I can build bijection between $\mathbb{R}$ and $\mathbb{R}$ ($[i,i+1) \mapsto (-i-1,-i]$).

Have no idea how to build such group. Any hints? It would be great if you're have some link about it.

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## 1 Answer

The only two groups for which $\mathrm{Aut}(G)$ is trivial are the trivial group $G = \{1\}$ and the cyclic group on two elements $G = \{\pm 1\}$.

To see this, first note that if $\mathrm{Aut}(G)$ is trivial, then $G$ is abelian, because if $h \in G$ is not central then $g \mapsto h^{-1}gh$ is a non-trivial automorphism. Now if $(G,+)$ (switching to additive notation) is an abelian group, then $g \mapsto -g$ is an automorphism (being its own inverse). If $\mathrm{Aut}(G)$ is trivial, this means that this automorphism is the identity, so $g = -g$ for each element $g \in G$.

In other words, all elements have order 2. This makes $G$ a vector space over the field $\mathbb F_2$ with two elements, simply by defining the multiplication with 0 to always give the identity element of $G$, and letting multiplication with 1 be the identity. Now, fix a basis $\{g_i\}_{i \in I}$ of $G$ as a vector space over $\mathbb F_2$.

If $|I| \geq 2$, you can define an automorphism of $G$ by exchanging $g_i$ and $g_j$, for some $i \neq j$, and leaving all other basis vectors fixed. Hence, $|I| \leq 1$, and this leaves us only with the two groups mentioned in the first sentence.