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
$$ \left(\begin{array}{cccc}10&6&0&0\\0&1&1&0\\0&10&0&0\\0&4&0&10\end{array}...
- 83.9k
4
votes
Accepted
Are exact sequence homomorphisms unique?
It's entirely possible to have multiple distinct short exact sequences. For a cheap example, you can take any (nontrivial) automorphism $\varphi$ of $B$. Then if we have a short exact sequence
$$
0 \...
- 31.4k
2
votes
Accepted
Let $G$ be a simple group. Show that any homomorphism from $ G$ to $G'$ (arbitrary $G'$) must be either injective or the trivial homomorphism.
If $G$ is a simple group, then by definition $G$ has only two normal subgroups: either $G$ or the trivial subgroup $\{e\}$. Now given any group homomorphism $f\colon G\to G'$ where $G$ is a simple ...
- 870
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
Can't understand matrix representation of a torsion subgroup homomorphism
In the text they say: Let $E$ be an elliptic curve over a field $K$ ... then
$$
E[n]\cong Z_n + Z_n
$$
The text gives no proof yet - they say it will come in section 3.2.
Let $n$ be such that the ...
- 48
1
vote
Accepted
If $f$ is a epimorphism from the monoid $(X,⊕,x_0)$ to the monoid $(Y,⊗, y_1)$ then $f(x_0)=y_1$ and $f(x^{-1})=f(x)^{-1}$ provided $x^{-1}$ exists.
Claim: Given a surjective semigroup homomorphism $f$ between two monoids $(X,\oplus,x_0),(Y,\otimes,y_1)$ (This means $f(x\oplus x') = f(x)\otimes f(x') $) then it is already a monoid homomorphism (...
- 1,077
1
vote
Accepted
Homomorphisms in the definition of free products
What you describe is the universal property of a coproduct. There is no explicit construction in your text, only the universal property, and depending on the construction the coproduct maps will be ...
- 3,066
1
vote
The Question "Prove that there is no homomorphism from Z16 ⊕ Z2 onto Z4 ⊕ Z4" In texbook, Is my proof correct?
I am not pure algebraist mathematician, but I thought an ideal solution would be in the following way:
If $φ$ is onto, ${\rm Ker}(φ)<\Bbb{Z}_{16}\oplus\Bbb Z_2$ is a normal subgroup of order $\frac{...
- 5,181
1
vote
Any permutation of {1,2,3,4} generates a permutation of the 6 subsets of {1,2,3,4} that have exactly 2 elements. So ...
The $24$ permutations used have the property that if two subsets share an element, the mappings of those two subsets also share an element.
Your assignment between six letters and the six pairs does ...
- 7,380
1
vote
Accepted
If $S = A\cdot A\cdot A$ generates $G$ and $f(S) = H$ is a group then $f(A) \leqslant H$ is also a group? Under what conditions is this true?
$\newcommand{\ZZ}{\mathbb Z}$ (Unless I'm misunderstanding the question) I don't think this is necessarily true. Suppose $G=\ZZ$, $H=\ZZ/3\ZZ$, $f$ is the quotient map, and $S=\{0,1,2,3\}$. Then $S$ ...
- 31.6k
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