Let $I$ be the $2\times 2$ identity matrix.
Then prove that $-I$ cannot be a commutator $[A, B]:=ABA^{-1}B^{-1}$ for any $2\times 2$ matrices $A$ and $B$ with determinant $1$.

Proof.

Assume that $[A, B]=-I$. Then $ABA^{-1}B^{-1}=-I$ implies
\[ABA^{-1}=-B. \tag{*}\] Taking the trace, we have
\[-\tr(B)=\tr(-B)=\tr(ABA^{-1})=tr(BAA^{-1})=\tr(B),\] hence the trace $\tr(B)=0$.
Thus, the characteristic polynomial of $B$ is
\[x^2-\tr(B)x+\det(B)=x^2+1.\] Hence the eigenvalues of $B$ are $\pm i$.

Note that the matrix $\begin{bmatrix}
0 & -1\\
1& 0
\end{bmatrix}$ has also eigenvalues $\pm i$.
Thus this matrix is similar to the matrix $B$ as both matrices are similar to the diagonal matrix $\begin{bmatrix}
i & 0\\
0& -i
\end{bmatrix}$.
Let $P$ be a nonsingular matrix such that
\[B’:=P^{-1}BP=\begin{bmatrix}
0 & -1\\
1& 0
\end{bmatrix}.\] Let $A’=P^{-1}AP$.

The relation (*) is equivalent to $AB=-BA$.
Using this we have
\begin{align*}
A’B’&=(P^{-1}AP)(P^{-1}BP)=P^{-1}(AB)P\\
&=P^{-1}(-BA)P=-(P^{-1}BP)(P^{-1}AP)=-B’A’.
\end{align*}

Let $A’=\begin{bmatrix}
a & b\\
c& d
\end{bmatrix}$.
Then $A’B’=-B’A’$ gives
\begin{align*}
\begin{bmatrix}
a & b\\
c& d
\end{bmatrix}
\begin{bmatrix}
0 & -1\\
1& 0
\end{bmatrix}
=-\begin{bmatrix}
0 & -1\\
1& 0
\end{bmatrix}\begin{bmatrix}
a & b\\
c& d
\end{bmatrix}\\[6pt] \Leftrightarrow
\begin{bmatrix}
b & -a\\
d& -c
\end{bmatrix}=\begin{bmatrix}
c & d\\
-a& -b
\end{bmatrix}.
\end{align*}
Hence we obtain $d=-a$ and $c=b$.

Then
\begin{align*}
1=\det(A)=\det(PA’P^{-1})=\det(A’)=\begin{vmatrix}
a & b\\
b& -a
\end{vmatrix}=-a^2-b^2,
\end{align*}
which is impossible.
Therefore, the matrix $-I$ cannot be written as a commutator $[A, B]$ for any $2\times 2$ matrices $A, B$ with determinant $1$.

The post An Example of a Matrix that Cannot Be a Commutator 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 $K, N$ be normal subgroups of a group $G$. Suppose that the quotient groups $G/K$ and $G/N$ are both abelian groups.

Then show that the group
\[G/(K \cap N)\] is also an abelian group.
 

Hint.

We use the following fact to prove the problem.

Lemma: For a subgroup $H$ of a group $G$, $H$ is normal in $G$ and $G/H$ is an abelian group if and only if the commutator subgroup $D(G)=[G,G]$ of $G$ is contained in $H$.

For a proof of this fact, see Commutator subgroup and abelian quotient group

Proof.

By the lemma mentioned above, we know that $G/K$ is an abelian group if and only if the commutator subgroup $D(G)=[G,G]$ is contained in $K$.

Similarly, since $G/N$ is abelian, $D(G)$ is contained in $N$.
Therefore, the commutator subgroup $D(G) \subset K \cap N$. This implies, again by Lemma, that the quotient group
\[G/(K \cap N)\] is an abelian group as required.

The post Two Quotients Groups are Abelian then Intersection Quotient is Abelian first appeared on Problems in Mathematics.

Let $G$ be a group and let $D(G)=[G,G]$ be the commutator subgroup of $G$.
Let $N$ be a subgroup of $G$.
Prove that the subgroup $N$ is normal in $G$ and $G/N$ is an abelian group if and only if $N \supset D(G)$.

Definitions.

Recall that for any $a, b \in G$, the commutator of $a$ and $b$ is
\[ [a,b]=a^{-1}b^{-1}ab \in G.\] The commutator subgroup $D(G)=[G,G]$ is a subgroup of $G$ generated by all commutators.
That is,
\[D(G)=[G,G]=\langle [a,b] \mid a,b \in G \rangle.\]

Proof.

$(\implies)$ Suppose that $N$ is a normal subgroup of $G$ and the quotient $G/N$ is an abelian group.
Then for any elements $a, b \in G$, we have
\begin{align*}
abN=aN\cdot bN=bN \cdot aN=baN.
\end{align*}
(Here we used the fact that $N$ is normal, hence $G/N$ is a group.)

From this, we obtain that
\[a^{-1}b^{-1}abN=N\] and thus $[a,b]=a^{-1}b^{-1}ab\in N$.
Since any generator $[a,b]$ is in $N$, we have $D(G)\subset N$.

$(\impliedby)$ On the other hand, let us assume that $N \supset D(G)$.
We first show that $N$ is a normal subgroup of $G$.
For any $g \in G$, $x\in N$, we have
\begin{align*}
gxg^{-1}=gxg^{-1}x^{-1}x=[g^{-1},x^{-1}]\cdot x\in N
\end{align*}
since the commutator $[g^{-1},x^{-1}]\in D(G)\subset N$ and $x \in N$.
Thus $N$ is normal in $G$.

Now that $N$ is normal in $G$, the quotient $G/N$ is a group. We show that $G/N$ is an abelian group.
For any $a, b \in G$, we have
\begin{align*}
aN\cdot bN&=ab N\\
&=baa^{-1}b^{-1}abN\\
&=ba[a,b]N\\
&=baN \qquad \text{ since } [a,b] \in N\\
&=bN\cdot aN.
\end{align*}
Therefore the group operations of $G/N$ is commutative, and hence $G/H$ is abelian.

Related Question.

For another abelian group problem, check out
Two quotients groups are abelian then intersection quotient is abelian

The post Commutator Subgroup and Abelian Quotient Group first appeared on Problems in Mathematics.

Let $G$ be a group and $H$ and $K$ be subgroups of $G$.
For $h \in H$, and $k \in K$, we define the commutator $[h, k]:=hkh^{-1}k^{-1}$.
Let $[H,K]$ be a subgroup of $G$ generated by all such commutators.

Show that if $H$ and $K$ are normal subgroups of $G$, then the subgroup $[H, K]$ is normal in $G$.

Proof.

We first prove that a conjugate of each generator is in $[H,K]$.
Let $h \in H, k \in K$. For any $g \in G$, we have by inserting $g^{-1}g=e$ inbetweens
\begin{align*}
g[h,k]g^{-1}&=ghkh^{-1}k^{-1}=(ghg^{-1})(gkg^{-1})(gh^{-1}g^{-1})(gk^{-1}g^{-1})\\
& = (ghg^{-1})(gkg^{-1})(ghg^{-1})^{-1}(gkg^{-1})^{-1}.
\end{align*}

Now note that $ghg^{-1} \in H$ since $H$ is normal in $G$, and $gkg^{-1} \in K$ since $K$ is normal in $G$. Thus $g[h,k]g^{-1} \in [H, K]$.
By taking the inverse of the above equality, we also see that $g[k,h]g^{-1} \in [H, K]$. Thus the conjugate of the inverse $[h,k]^{-1}=[k,h]$ is in $[H, K]$.

Next, note that any element $x \in [H,K]$ is a product of generators or their inverses. So let us write
\[x=[h_1, k_1]^{\pm 1}[h_2, k_2]^{\pm 1}\cdots [h_n, k_n]^{\pm 1},\] where $h_i \in H, k_i\in K$ for $i=1,\dots, n$.
Then for any $g \in G$, we have
\begin{align*}
gxg^{-1}=(g[h_1, k_1]^{\pm 1}g^{-1})(g[h_2, k_2]^{\pm 1}g^{-1})\cdots(g [h_n, k_n]^{\pm 1} g^{-1}).
\end{align*}

We saw that the conjugate of a generator, or its inverse, by $g \in G$ is in $[H,K]$.
Thus $gxg^{-1}$ is also in $[H, K]$.
This proves that the group $[H,K]$ is a normal subgroup of $G$.

Related Question.

Another problem about a commutator group is
A condition that a commutator group is a normal subgroup

The post Group Generated by Commutators of Two Normal Subgroups is a Normal Subgroup first appeared on Problems in Mathematics.

Let $H$ be a normal subgroup of a group $G$.
Then show that $N:=[H, G]$ is a subgroup of $H$ and $N \triangleleft G$.

Here $[H, G]$ is a subgroup of $G$ generated by commutators $[h,k]:=hkh^{-1}k^{-1}$.

In particular, the commutator subgroup $[G, G]$ is a normal subgroup of $G$

Proof.

First, we show that $N=[H, G]$ is a subgroup of $H$.
A generator of $N$ is of the form either $hgh^{-1} g^{-1}$ or $ghg^{-1} h^{-1}$, where $h \in H$ and $g \in G$. Since $H$ is normal in $G$, we see that these are elements in $H$. Thus $N <H$.

Next, we show that $N$ is normal in $G$.

Let $x=hgh^{-1} g^{-1}$ be a generator element of $N$.

Then for any $a \in G$, we have
\[ ax a^{-1} =ahgh^{-1} g^{-1}a^{-1}=(aha^{-1})(aga^{-1})(ah^{-1}a^{-1})(a g^{-1}a^{-1}) \in [H, G]. \] Similarly, if $x$ is a generator of the form $ghg^{-1} h^{-1}$, then we see that $ghg^{-1} h^{-1} \in [H, G]$ by the same argument.
Thus the conjugate of a generator of $N$ by $a \in G$ stays in $N$.

Now any element $x \in N$ is of the form $x=x_1 x_2 \cdots x_n$, where $x_i$ are generators of $N$.

For any $a \in H$, we have
\[ axa^{-1}=a x_1 x_2 \cdots x_n a^{-1}= (ax_1 a^{-1})(a x_2 a^{-1}) \cdots (a x_n a^{-1}) \in [H, G].\] Thus $N \triangleleft G$.

In particular, we apply the result to $H=G$. Then we see that the commutator subgroup $[G, G]$ is a normal subgroup of $G$.

Related Question.

You might also be interested in the problems:

• A condition that a commutator group is a normal subgroup
• Non-abelian simple group is equal to its commutator subgroup

The post A Condition that a Commutator Group is a Normal Subgroup first appeared on Problems in Mathematics.