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### 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
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### 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
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### 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
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### 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
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### 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

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\$ ...
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