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
$$ \left(\begin{array}{cccc}10&6&0&0\\0&1&1&0\\0&10&0&0\\0&4&0&10\end{array}...
- 83.9k
3
votes
Does every extension of a finite group by $\mathbb{R}^n$ split?
This is a standard cocycle averaging situation.
Choose an arbitrary (set-theoretic) section $s : G/N \to G$. The extent to which $s(G/N)$ fails to be a subgroup is measured by the cocycle $f : (G/N)^2 ...
- 7,671
2
votes
Why isn’t every subgroup of a Finite Group Cyclic?
Take a non-cyclic group $G$ and any group $H$. Then, $K:=G\times\{1_H\}\cong G$ and $K\le G\times H$.
- 1,447
2
votes
Accepted
Is this proof that a homomorphism preserves identities correct (sufficient)?
Because every element in a group has an inverse, it's enough for
$$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 $yz^{-1}$ ...
- 8,986
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