1
$\begingroup$

(38.) $\mathbb{Z} \rightarrow S_3$?

Let $φ(n) = \begin{cases} \mathrm{id} \in S_3 &, \text{for all $n$ even,} \\ \mathrm{transposition} (1,2) &, \text{for all $n$ odd integers.} \end{cases}$
Note that (1, 2) is of order 2, isomorphic to Z2.

(41.) $D_4 \rightarrow S_3$?

View D4 as a group of permutations. Same answer as (43.) underneath to $D_4$, just change $S_4$ to $D_4$.

(43.) $S_4 \rightarrow S_3$?

Viewing D4 as a group of permutations, let $φ(p) = \begin{cases} \mathrm{id} \in S_4 &, \text{for all $p$ even permutations,} \\ (1,2) &, \text{for all $p$ odd permutations.} \end{cases}$ Note that (1, 2) is a subgroup of S3 of order 2, isomorphic to Z2.

(1.) I see the image is $S_3$ every time. However I don't understand why the same homomorphism works in all three questions? What's the connection between them? What's the intuition?

(2.) How do you magically envisage and envision this hard piecewise-defined homomorphism?

(3.) What other homomorphisms work for all three? How many are there?

$\endgroup$
0
$\begingroup$

The common thread is that even+even = odd+odd = even, and even+odd = odd+even = odd; and this is the structure of $\Bbb Z_2$ (as they mention each time). So any group that has a notion of odd/even that satisfies the above will have this sort of "reduction to $\Bbb Z_2$" that has been demonstrated here.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.