Talk:Abstract Algebra/Group Theory/Homomorphism

Ummm. It seems to me that, contrary to the statement of the exercise at the bottom of the page, there is a non-trivial homomorphism from $$\mathbb{Z}_5$$ to $$S_3$$.

First, let $$z$$ be a generator of $$\mathbb{Z}_5$$, so that the remaining elements of $$\mathbb{Z}_5$$ are the identity, $$e$$, and $$z^2$$, $$z^3$$, and $$z^4$$. Next, consider $$S_3$$ as the group of permutations of the letters $$a$$, $$b$$, and $$c$$, and, in particular, we consider $$t$$, that permutation of $$a$$, $$b$$, and $$c$$ that maps the ordered triple $$(a, b, c)$$ to the triple $$(b, a, c)$$, i.e $$t$$ is the transposition of $$a$$ and $$b$$.

Now, consider the mapping $$m$$ from $$\mathbb{Z}_5$$ to $$S_3$$ that takes $$z^k$$ to the identity, $$i$$, of $$S_3$$ if $$k$$ is even, and takes the element $$z^k$$ to $$t$$ if $$k$$ is odd. Since all elements of $$\mathbb{Z}_5$$ can be written as zk for some integer $$k$$, $$m$$ is well defined. Furthermore, it is evident that $$m(z_1 z_2) = m(z_1) m(z_2)$$ for each $$z_1, z_2$$ in $$\mathbb{Z}_5$$.

$$m$$ is therefore a homomorphism.

So maybe it would be better for the exercise to read "find a a non-trivial homomorphism from $$\mathbb{Z}_5$$ to $$S_3$$". There are, of course, 3 such homomorphisms, as the transpositions of letters $$a$$, and $$c$$ or $$b$$, and $$c$$ will also work. User:Eweinber|Eweinber]] (discuss • contribs) 20:57, 15 October 2017 (UTC)