Question

Give an example of a nontrivial homomorphism for the given groups, if an example exist. If no such homomorphism exist, explain why that is so. ϕ:D4-->S3

Answer 1

let us denote the dihedral group by \(D_4=\{e,R,R^2,R^3,F,FR,FR^2,FR^3\}\).

Now we have that \(R,F\) are the generators, we must send \(f(F)\) to something that divides the order of \(F\).

\(S_3=\{I, (12),(23),(13),(123),(132)\}\).

Since \(F\) has order \(2\) we only have \(4\) (namely \(I,(12),(23),(13)\)) options, so lets choose one of them. Say \((12)\). Then \(f(F)=(12)\).

Similarly \(R\) has order \(4\) so we have the same options to send it to.(because 1,2 both divide 4, but 3 does not)

Lets choose what element to send \(R\) to. Say \((23)\), so \(f(R)=(23)\).

Now,

\(f(R)=(12)\\f(R^2)=f(R)^2=(12)^2=(12)(12)=I\\f(R^3)=f(R)^3=(12)^3=(12)(12)(12)=(12)\\


\\f(F)=(23)\\
f(FR)=f(F)f(R)=(23)(12)=(132)\\
f(FR^2)=f(F)f(R)f(R)=(23)(12)(12)=(23)\\
f(FR^3)=f(F)f(R)f(R)f(R)=(23)(12)(12)(12)=(23)(12)=(132)
\\
f(e)=f(F^2)=f(F)^2=(23)^2=(23)(23)=I\)

\(f\) is a homomorphism.

We must check to see if our generators commute (why?)

\((12)(23)=(23)(12) \ \ \ \ \ \checkmark\)

Note that you can find all the homomorphism's by matching up generators with the other elements I listed.

Answer 2

Thank you so much! I appericate it!

Answer 3

How did you figure out that F has an order of 2?

Answer 4

think about that F is, it flips the square across some axis. If we do it twice, we are back to where we started. Flips will always have order 2, in dihedral groups.

Answer 5

If you flip it twice on any of those lines, you get back to where you started