7 votes
Accepted

A non-trivial homomorphism from $SL(2, 7)$ to $GL(4, 11)$.

Yes, there is an injective homomorphism with image generated by the matrices $$ \left(\begin{array}{cccc}10&6&0&0\\0&1&1&0\\0&10&0&0\\0&4&0&10\end{array}...
  • 83.9k
3 votes

Does every extension of a finite group by $\mathbb{R}^n$ split?

This is a standard cocycle averaging situation. Choose an arbitrary (set-theoretic) section $s : G/N \to G$. The extent to which $s(G/N)$ fails to be a subgroup is measured by the cocycle $f : (G/N)^2 ...
2 votes

Why isn’t every subgroup of a Finite Group Cyclic?

Take a non-cyclic group $G$ and any group $H$. Then, $K:=G\times\{1_H\}\cong G$ and $K\le G\times H$.
  • 1,447
2 votes
Accepted

Is this proof that a homomorphism preserves identities correct (sufficient)?

Because every element in a group has an inverse, it's enough for $$fg' = g'$$ for just one $g'\in G'$ to prove that $f$ is the identity. The fact that we already know that $G'$ is a group has done ...
1 vote
Accepted

A ternary relation on a group

Start with your $x y^{-1} z x^{-1} y z^{-1} = 1$. Multiply both sides by $z$ from the right, then multiply both sides by $z^{-1}$ from the left. You get $z^{-1} x y^{-1} z x^{-1} y = 1$. In this way ...
  • 40.5k
1 vote

A ternary relation on a group

Since $1=1^{-1}$ we have $(x y^{-1} z x^{-1} y z^{-1})^{-1} = zy^{-1}xz^{-1}yx^{-1} = 1$ which gives us $R(x,y,z)=R(z,y,x)$. With $x y^{-1} z x^{-1} y z^{-1} = 1$ we multiply on the left by $yz^{-1}$ ...

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