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Shaun
Shaun
2:54 PM
@UnknownMathMan, a group isomorphism is defined the way it is because, in a sense, the axioms of a group are invariant under the isomorphism; that is, for an iso $\phi: G\to H$, we have that: $(\phi(a)\phi(b))\phi(c)=\phi(a)(\phi(b)\phi(c))$ for all $a,b,c\in G$; if $e$ is the identity of $G$, then $\phi(e)$ is the identity of $H$; and for all $g\in G$, we have $\phi(g^{-1})=(\phi(g))^{-1}$. The fact that the iso is a bijection means that the same number of elements are in each group.
@UnknownMathMan, in fact, I'm not so sure of that explanation . . .
 
Shaun
Shaun
3:27 PM
Quick question:

Let $\phi: G\to H$ be a surjective homomorphism. Suppose that $H$ is virtually solvable. Is $G$ then virtually solvable?
 
Shaun
Shaun
3:58 PM
in Mathematics, 1 min ago, by Shaun
Well, if $G$ has a free, non-abelian subgroup, then $G$ cannot be virtually solvable, but $G$ is homomorphic to the trivial group, which is virtually solvable.
 
 
2 hours later…
Buddhini Angelika
Buddhini Angelika
5:58 PM
Hi, if a cayley graph is hamiltonian connected we can directly say that the graph is hamiltonian, right?
 
 
5 hours later…
Shaun
Shaun
11:17 PM
Yes, @BuddhiniAngelika
 

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