I want to prove that every finite group $G$ of order more than 2 has a nontrivial automorphism.  I've seen this question answered on this site for infinite groups, but the proofs given use the fact that if $g^2=1$ for every $g$ in $G$, then $G$ is a vector space over $\mathbb{Z}_2$.  This is an exercise in Herstein's text that appears before the section on (the fundamental theorem of) finite abelian groups.  I think I can prove this result using that theorem, but was wondering if there are more elementary proofs.  Here is my proof: If $G$ is nonabelian, then $\exists x \in G$ such that the map  $(T_x: g \mapsto x^{-1}gx)$ is a nontrivial automorphism of $G$.  So suppose $G$ is abelian.  Then the map $g \mapsto g^{-1}$ is an automorphism of $G$; this automorphism is nontrivial if some element in $G$ has order at least 3.  If every element in $G$ has order 2, then by the fundamental theorem on finite abelian groups, $G \cong C_2 \times \cdots \times C_2$ is the direct product of $k$ copies of $C_2$ for some $k \ge 2$.  A map that interchanges the generators of the first two copies and fixes the remaining $k-2$ copies yields a nontrivial automorphism of $G$. QED.
Well, this is an exercise problem from Herstein which sounds difficult:   How does one prove that if $|G|&gt;2$, then $G$ has non-trivial automorphism?   The only thing I know which connects a group with its automorphism is the theorem, $$G/Z(G) \cong \mathcal{I}(G)$$ where $\mathcal{I}(G)$ denotes the Inner- Automorphism group of $G$. So for a group with $Z(G)=(e)$, we can conclude that it has a non-trivial automorphism, but what about groups with center?
Drop or change automatic migration-rejection based on non-existing tags I believe this feature creates more issues than it solves and thus would like to see it gone. (This is an out-growth of , which I repost here as it concerns several/all sites.)  The main reason why I think so is that there can be numerous reasons other than the subject being off-topic why some tag does not exist on some site, for one thing it could be a spelling, typesetting, or other formalistic reason. For example, for the two sites I know best MathOverflow and Mathematics hardly any of the most popular tags have a strictly matching tag on Mathematics but only because on MO there is the convention that certain tags are formatted in a specific way, which is not in place on Mathematics (for example nt.number-theory vs number-theory).  On top of that a user choosing the wrong site in the first place is likely not too well aware of the tags of the site they are posting too so that the question might be poorly tagged in the first place.         There is an additional detail I do not understand (it is obsolete if the actual request should be followed but in case it is not):   Even if on the source site a tag has a synonym and that synonym exists as a tag on the target site the migration is rejected. I do not understand the logic behind this at all. Possibly this is just an oversight.   I could also think of some other ways to improve the process while keeping the idea, but since I believe that it is best dropped I do not (yet) elaborate on this. Is it possible to prevent LUDecomposition from pivoting? What does “{} \;” mean in find (in Linux)? Sometimes I see a command like  find . -name * -exec ls -a {} \;   I was asked to execute this.  What does {} \; mean here? 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Find all continuous functions $f$ satisfying   $$\int_0^xf=(f(x))^2+C$$ for some constant $C \neq 0$, assuming that $f$ has at most one $0$.  I have a question about the solution to this problem. It says that   Clearly $f^2$ is differentiable everywhere, its derivative at $x$ is $f(x)$ (I'm assuming that this is because the left hand side is differentiable, the right hand side must be differentiable as well). So $f$ is differentiable at $x$ whenever $f(x) \neq 0$, and  $$f(x)=2f(x)f'(x),$$ so $f'(x) = \frac{1}{2}$ at such points. Thus, $f= \frac{1}{2}x+b$ for some $b$ on any interval where it is non-zero; if $f$ has a zero, this gives two possible solutions, with two possible values of $b$, but since $f$ is assumed continuous, they must be the same. So we need $$\int_0^x(\frac{1}{2}t+b)dt=(\frac{1}{2}x+b)^2+C$$  so we must have $b=\sqrt{-C}$ or $-\sqrt{-C}$, leading to two solutions for $C \lt 0$.  I don't understand the bolded statements. First, how does the differentiability of $f^2$ guarantee that $f$ is differentiable, and why only at $x$ where $f(x) \neq 0$?  Next, why are there two possible solutions if $f$ has a zero, but since $f$ is continuous they must be the same?  Finally, does the last statement mean that if $C \gt 0$, then there is no solution, since we're restricted to real numbers here?  I would appreciate it if anyone answers my questions. What is ISCAS85 benchmark? What is ISCAS85 benchmark? I find it in a , is there any simulator for it? Level 10 vim adventures beers on the wall complete solution I saw this question  but it doesn't have a full solution. What is the full solution to the 99 bottles of beer on the wall for Vim adventures?  Context: for the sake of being a good SO member I will give more details but a full solution will answer my question. It seems from reading the other post that my register is incorrect for some reason. I can't get the &lt;NL&gt;. 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