Let $G$ be a group. (Do not assume that $G$ is a finite group.)
Prove that $G$ is a simple abelian group if and only if the order of $G$ is a prime number.

Definition.

A group $G$ is called simple if $G$ is a nontrivial group and the only normal subgroups of $G$ is either the trivial group or $G$ itself.

Proof.

$(\implies)$ If $G$ is a simple abelian group, then the order of $G$ is prime.

Suppose that $G$ is a simple abelian group. Then $G$ is a nontrivial group by definition.

We first show that $G$ is a finite group.
Let $g\in G$ be a nonidentity element of $G$. Then the group $\langle g \rangle$ generated by $g$ is a subgroup of $G$. Since $G$ is an abelian group, every subgroup is a normal subgroup.

Since $G$ is simple, we must have $\langle g \rangle=G$. If the order of $g$ is not finite, then $\langle g^2 \rangle$ is a proper normal subgroup of $\langle g \rangle=G$, which is impossible since $G$ is simple.
Thus the order of $g$ is finite, and hence $G=\langle g \rangle$ is a finite group.

Let $p$ be the order of $g$ (hence the order of $G$).
Seeking a contradiction, assume that $p=mn$ is a composite number with integers $m>1, n>1$. Then $\langle g^m \rangle$ is a proper normal subgroup of $G$. This is a contradiction since $G$ is simple.

Thus $p$ must be a prime number.
Therefore, the order of $G$ is a prime number.

$(\impliedby)$ If the order of $G$ is prime, then $G$ is a simple abelian group.

Let us now suppose that the order of $G$ is a prime.

Let $g\in G$ be a nonidentity element. Then the order of the subgroup $\langle g \rangle$ must be a divisor of the order of $G$, hence it must be $p$.

Therefore we have $G=\langle g \rangle$, and $G$ is a cyclic group and in particular an abelian group.
Since any normal subgroup $H$ of $G$ has order $1$ or $p$, $H$ must be either trivial $\{e\}$ or $G$ itself. Hence $G$ is simple. Thus, $G$ is a simple abelian group.

The post A Simple Abelian Group if and only if the Order is a Prime Number first appeared on Problems in Mathematics.

Let $G$ be a non-abelian simple group. Let $D(G)=[G,G]$ be the commutator subgroup of $G$. Show that $G=D(G)$.
 

Definitions/Hint.

We first recall relevant definitions.

• A group is called simple if its normal subgroups are either the trivial subgroup or the group itself.
• The commutator subgroup $D(G)=[G,G]$ is a subgroup of $G$ generated by all commutators $[a,b]=a^{-1}b^{-1}ab$ for $a,b\in G$.

The commutator subgroup $D(G)=[G,G]$ is a normal subgroup of $G$.
For a proof, see: A condition that a commutator group is a normal subgroup.

Proof.

Note that the commutator subgroup $D(G)$ is a normal subgroup.
Since $G$ is simple, any normal subgroup of $G$ is either the trivial group $\{e\}$ or $G$ itself. Thus we have either $D(G)=\{e\}$ or $D(G)=G$.
If $D(G)=\{e\}$, then for any two elements $a,b \in G$ the commutator $[a,b]\in D(G)=\{e\}$.

Thus we have
\[a^{-1}b^{-1}ab=[a,b]=e.\] Therefore we have $ab=ba$ for any $a,b\in G$. This means that the group $G$ is abelian, which contradicts with the assumption that $G$ is non-abelian.
Therefore, we must have $D(G)=G$ as required.

The post Non-Abelian Simple Group is Equal to its Commutator Subgroup first appeared on Problems in Mathematics.

Let $G$ be a finite group. Then show that $G$ has a composition series.

Proof.

We prove the statement by induction on the order $|G|=n$ of the finite group.
When $n=1$, this is trivial.

Suppose that any finite group of order less than $n$ has a composition series.
Let $G$ be a finite group of order $n$.
If $G$ is simple, then $G \rhd \{e\}$, where $e$ is the identity element of $G$, is a composition series and we are done.

Thus, suppose that $G$ is not simple. Then it has a nontrivial proper normal subgroup.
Since $G$ is a finite group, there exists a maximal proper normal subgroup $H$.

Then the quotient $G/H$ is a simple group.
In fact, if $N$ is a proper normal subgroup of $G/H$, then the preimage of $N$ under the natural projection homomorphism $\pi:G \to G/H$ is a proper normal subgroup of $G$ containing $H$ by the fourth isomorphism theorem.
Since $H$ is maximal, the preimage $\pi^{-1}(N)$ must be $H$. This implies $N$ is trivial in $G/H$ and thus $G/H$ is simple.

Since $H$ is a proper subgroup of $G$, the order of $H$ is less than that of $G$.
Thus by the induction hypothesis, $H$ has a composition series
\[H=N_1 \rhd N_1 \rhd \cdots \rhd N_m=\{e\}.\]

The series
\[G=N_0 \rhd H=N_1 \rhd N_1 \rhd \cdots \rhd N_m=\{e\}\] has a simple factors $N_i/N_{i+i}$, hence it is a composition series for $G$.

Related Question.

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Infinite cyclic groups do not have composition series

The post Any Finite Group Has a Composition Series first appeared on Problems in Mathematics.

Let $G$ be a simple group and let $X$ be a finite set.
Suppose $G$ acts nontrivially on $X$. That is, there exist $g\in G$ and $x \in X$ such that $g\cdot x \neq x$.
Then show that $G$ is a finite group and the order of $G$ divides $|X|!$.

Proof.

Since $G$ acts on $X$, it induces a permutation representation
\[\rho: G \to S_{X}.\]

Let $N=\ker(\rho)$ be the kernel of $\rho$.
Since a kernel is normal in $G$ and $G$ is simple, we have either $N=\{e\}$ or $N=G$.

If $N=G$, then for any $g\in G$ we have $\rho(g)$ is a trivial action, that is, $g\cdot x=x$ for any $X$.
This contradicts the assumption that $G$ acts nontrivially on $X$.
Hence we have $N=\{e\}$, and it follows that the homomorphism $\rho$ is injective.

Thus we have
\[G \cong \mathrm{im} (\rho) < S_{X}.\] Since $S_{X}$ is a finite group and $G$ is isomorphic to its subgroup, the group $G$ is finite.
By Lagrange’s theorem, the order $|G|=|\mathrm{im}(\rho)|$ of $G$ divides the order $|S_{X}|=|X|!$ of $S_{X}$.

The post Nontrivial Action of a Simple Group on a Finite Set first appeared on Problems in Mathematics.

(a) Show that if a group $G$ has the following order, then it is not simple.

1. $28$
2. $496$
3. $8128$

(b) Show that if the order of a group $G$ is equal to an even perfect number then the group is not simple.

Hint.

Use Sylow’s theorem.
(See the post Sylow’s Theorem (summary) to review the theorem.)

Proof.

(a) A group of the following order is not simple

(1) A group of order $28$

Note that $28=2^2\cdot 7$. The number $n_7$ of the Sylow $7$-subgroups of $G$ satisfies
\[n_7 \equiv 1 \pmod{7} \text{ and } n_7|2^2.\] Thus, the only possible value is $n_7=1$. The unique Sylow $7$-subgroup is a proper nontrivial normal subgroup of $G$, hence $G$ is not simple.

(2) A group of order $496$

Note that $496=2^4\cdot 31$. By the same argument as in (1), there is a normal Sylow $31$-subgroup in $G$, hence $G$ is not simple.

(3) A group of order $8128$

We have $8128=2^6\cdot 127$, where $127$ is a prime number.
Again the same reasoning proves that the group $G$ has the unique normal Sylow $127$-subgroup in $G$, hence $G$ is not simple.

(b) If the order is an even perfect number, a group is not simple

From elementary number theory, all even perfect numbers are of the form
\[2^{p-1}(2^p-1),\] where $p$ is a prime number and $2^p-1$ is also a prime number.
(For a proof, see the post “Even Perfect Numbers and Mersenne Prime Numbers“.)

Suppose the order of a group $G$ is $2^{p-1}(2^p-1)$, with prime $p$, $2^p-1$.
Then the number $n_{2^p-1}$ of Sylow $(2^p-1)$-subgroup satisfies
\[n_{2^p-1}\equiv 1 \pmod {2^p-1} \text{ and } n_{2^p-1}|2^{p-1}.\] These force that $n_{2^p-1}=1$.
Therefore the group $G$ contains the unique normal Sylow $(2^p-1)$-subgroup, hence $G$ is not simple.

Similar problem

For an analogous problem, check out: Groups of order 100, 200. Is it simple?

Comment.

In about 300 BC Euclid showed that a number of the form $2^{p-1}(2^p-1)$ where $p$ and $2^p-1$ are prime numbers.
(If $2^p-1$ is a prime number, then $p$ must be a prime.)
The converse was proved by Euler in the 18th century. Namely, Euler proved that any even perfect number is of the form $2^{p-1}(2^p-1)$ with prime $2^p-1$.

The number of the form $2^p-1$ is called a Mersenne number and if it is a prime number, then it is called a Mersenne Prime.
It is unknown whether there are infinitely many Mersenne prime numbers.

Also it is unknown whether there is an odd perfect number.

The post If the Order is an Even Perfect Number, then a Group is not Simple first appeared on Problems in Mathematics.