Every Finite Group Having More than Two Elements Has a Nontrivial Automorphism

Michigan State University Abstract Algebra Exam Problems and Solutions

Problem 495

Prove that every finite group having more than two elements has a nontrivial automorphism.

(Michigan State University, Abstract Algebra Qualifying Exam)
 
LoadingAdd to solve later

Sponsored Links


Proof.

Let $G$ be a finite group and $|G|> 2$.

Case When $G$ is a Non-Abelian Group

Let us first consider the case when $G$ is a non-abelian group.
Then there exist elements $g, h\in G$ such that $gh\neq hg$.

Consider the map $\phi: G \to G$ defined by sending $x\in G$ to $gxg^{-1}$.
Then it is straightforward to check that $\phi$ is a group homomorphism and its inverse is given by the conjugation by $g^{-1}$.
Hence $\phi$ is an automorphism.

If $\phi=1$, then we have $h=\phi(h)=ghg^{-1}$, and this implies that $gh=hg$.
This contradicts our choice of $g$ and $h$.
Hence $\phi$ is a non-trivial automorphism of $G$.

Case When $G$ is an Abelian Group

Next consider the case when $G$ is a finite abelian group of order greater than $2$.
Since $G$ is an abelian group the map $\psi:G\to G$ given by $x \mapsto x^{-1}$ is an isomorphism, hence an automorphism.

If $\psi$ is a trivial automorphism, then we have $x=\psi(x)=x^{-1}$.
Thus, $x^2=e$, where $e$ is the identity element of $G$.

Sub-Case When $G$ has an Element of Order $\geq 3$.

Therefore, if $G$ has at least one element of order greater than $2$, then $\psi$ is a non-trivial automorphism.

Sub-Case When Elements of $G$ has order $\leq 2$.

It remains to consider the case when $G$ is a finite abelian group such that $x^2=e$ for all elements $x\in G$.
In this case, the group $G$ is isomorphic to
\[\Zmod{2}\times \Zmod{2}\times \cdots \Zmod{2}=(\Zmod{2})^n.\] Since $|G| > 2$, we have $n>1$.

Then the map $(\Zmod{2})^n\to (\Zmod{2})^n$ defined by exchanging the first two entries
\[(x_1, x_2, x_3, \dots, x_n) \mapsto (x_2, x_1, x_3, \dots, x_n)\] is an example of nontrivial automorphism of $G$.

Therefore, in any case, the group $G$ has a nontrivial automorphism.


LoadingAdd to solve later

Sponsored Links

More from my site

  • A Subgroup of Index a Prime $p$ of a Group of Order $p^n$ is NormalA Subgroup of Index a Prime $p$ of a Group of Order $p^n$ is Normal Let $G$ be a finite group of order $p^n$, where $p$ is a prime number and $n$ is a positive integer. Suppose that $H$ is a subgroup of $G$ with index $[G:P]=p$. Then prove that $H$ is a normal subgroup of $G$. (Michigan State University, Abstract Algebra Qualifying […]
  • Isomorphism Criterion of Semidirect Product of GroupsIsomorphism Criterion of Semidirect Product of Groups Let $A$, $B$ be groups. Let $\phi:B \to \Aut(A)$ be a group homomorphism. The semidirect product $A \rtimes_{\phi} B$ with respect to $\phi$ is a group whose underlying set is $A \times B$ with group operation \[(a_1, b_1)\cdot (a_2, b_2)=(a_1\phi(b_1)(a_2), b_1b_2),\] where $a_i […]
  • If Squares of Elements in a Group Lie in a Subgroup, then It is a Normal SubgroupIf Squares of Elements in a Group Lie in a Subgroup, then It is a Normal Subgroup Let $H$ be a subgroup of a group $G$. Suppose that for each element $x\in G$, we have $x^2\in H$. Then prove that $H$ is a normal subgroup of $G$. (Purdue University, Abstract Algebra Qualifying Exam)   Proof. To show that $H$ is a normal subgroup of […]
  • Abelian Normal subgroup, Quotient Group, and Automorphism GroupAbelian Normal subgroup, Quotient Group, and Automorphism Group Let $G$ be a finite group and let $N$ be a normal abelian subgroup of $G$. Let $\Aut(N)$ be the group of automorphisms of $G$. Suppose that the orders of groups $G/N$ and $\Aut(N)$ are relatively prime. Then prove that $N$ is contained in the center of […]
  • The Existence of an Element in an Abelian Group of Order the Least Common Multiple of Two ElementsThe Existence of an Element in an Abelian Group of Order the Least Common Multiple of Two Elements Let $G$ be an abelian group. Let $a$ and $b$ be elements in $G$ of order $m$ and $n$, respectively. Prove that there exists an element $c$ in $G$ such that the order of $c$ is the least common multiple of $m$ and $n$. Also determine whether the statement is true if $G$ is a […]
  • Non-Abelian Group of Order $pq$ and its Sylow SubgroupsNon-Abelian Group of Order $pq$ and its Sylow Subgroups Let $G$ be a non-abelian group of order $pq$, where $p, q$ are prime numbers satisfying $q \equiv 1 \pmod p$. Prove that a $q$-Sylow subgroup of $G$ is normal and the number of $p$-Sylow subgroups are $q$.   Hint. Use Sylow's theorem. To review Sylow's theorem, check […]
  • The Order of $ab$ and $ba$ in a Group are the SameThe Order of $ab$ and $ba$ in a Group are the Same Let $G$ be a finite group. Let $a, b$ be elements of $G$. Prove that the order of $ab$ is equal to the order of $ba$. (Of course do not assume that $G$ is an abelian group.)   Proof. Let $n$ and $m$ be the order of $ab$ and $ba$, respectively. That is, \[(ab)^n=e, […]
  • Order of Product of Two Elements in a GroupOrder of Product of Two Elements in a Group Let $G$ be a group. Let $a$ and $b$ be elements of $G$. If the order of $a, b$ are $m, n$ respectively, then is it true that the order of the product $ab$ divides $mn$? If so give a proof. If not, give a counterexample.   Proof. We claim that it is not true. As a […]

You may also like...

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.

More in Group Theory
Group Theory Problems and Solutions in Mathematics
If Two Subsets $A, B$ of a Finite Group $G$ are Large Enough, then $G=AB$

Let $G$ be a finite group and let $A, B$ be subsets of $G$ satisfying \[|A|+|B| > |G|.\] Here $|X|$...

Close