## An Example of a Matrix that Cannot Be a Commutator

## Problem 565

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 $1$.

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 $1$.

Let $G$ be a group. Suppose that we have

\[(ab)^3=a^3b^3\]
for any elements $a, b$ in $G$. Also suppose that $G$ has no elements of order $3$.

Then prove that $G$ is an abelian group.

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

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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.

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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 $N \supset D(G)$.

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 $G$.

Add to solve laterLet $A$ and $B$ be normal subgroups of a group $G$. Suppose $A\cap B=\{e\}$, where $e$ is the unit element of the group $G$.

Show that for any $a \in A$ and $b \in B$ we have $ab=ba$.

Let $A$ and $B$ be $n\times n$ matrices.

Suppose that these matrices have a common eigenvector $\mathbf{x}$.

Show that $\det(AB-BA)=0$.

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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 $G$

Add to solve later