# Abelian/Isomorphic logic statement

I don't understand this logic statement. I don't think the context helps at all but I thought i'd include it anyway.

$G$ is abelian and $H$ is not abelian, then $G\ncong H$, is the same as:

$G$ abelian and $G\cong H$, then $H$ is abelian

• A and B $\implies$ C is the same as A and not C $\implies$ not B – Mathmo123 Jun 29 '14 at 13:30

## 2 Answers

Context is irrelevant.

"If $G$ is abelian and $H$ is not abelian, then $G\ncong H$" has the form $(A\land \neg B)\to \neg C$.

And since $(A\land \neg B)\to \neg C\iff \neg A\lor B\lor \neg C\iff (A\land C)\to B$,

for appropriate interpretations of $A, B$ and $C$, you get what you want.

Let $P \equiv H\ \text{is not Abelian}"$ and $Q \equiv G \not\cong H"$.

Assume that $G$ is Abelian. Then the first statement says $P \Rightarrow Q$, and the second says $\sim Q \Rightarrow \sim P$. Of course, these two are equivalent.

Only, that's not exactly what's given, I simplified it. Let $A \equiv G\ \text{is Abelian}"$. What is given is $(A \wedge P) \Rightarrow Q$. This is the same as $(A \wedge \sim Q) \Rightarrow \sim P$, in exactly the same way as the simpler versions above.