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ThePerfectHacker · #3 December 18th, 2006, 05:50 PM

Quote:

Originally Posted by galactus

This is a homomorphism. If $x^{2}=1$ for [math]x\neq{1}

Again, I did not select problems that require anything that is even slightly advanced, like homomorphisms. But I do not see how, $\phi: G\to G$ defined as squaring the element is a homomorphism. Because we require that $\phi(xy)=\phi(x)\phi(y)$ not $\phi(xy)=\phi(yx)$ . If the problem was $(ab)^2=a^2b^2$ then yes that would be a homomorphism. But then the solution is simple expand $abab=aabb$ cancelation laws state that $ab=ba$ . Q.E.D.

Which textbook, you use?

Funny, I was with my advisor today registering for math classes. He was speaking on the phone with somebody telling that he will teach an algebra class. He mentioned he is going to use "my book". I wonder if he can to that conclusion before I mentioned it

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