hint: by the first isomorphism theorem, any homomorphic image of D4 is isomorphic to a quotient group of D4.
so, first, find all the normal subgroups of D4. if memory serves me right, there are only 4 of them.
hint: by the first isomorphism theorem, any homomorphic image of D4 is isomorphic to a quotient group of D4.
so, first, find all the normal subgroups of D4. if memory serves me right, there are only 4 of them.
call the subgroups H,K,M and N (we should also probably include the trivial normal subgroups {1} and D4).
find D4/D4 ≅ {1}, D4/H, D4/K, D4/M, D4/N and D4/{1} = D4.
here's what you know: K,L, and M are of order 4, so D4/K, D4/M and D4/N are of order 2. that makes describing D4/K, D4/M and D4/N easy.
so the only really "interesting example" is D4/H, which is of order 4. there are only 2 group types of order 4.
calculate D4/H. i'll get you started:
H = {1, r^2}. so here are the 4 cosets:
H
Hr = {r,r^3}
Hs = {s, r^2s} = {s,sr^2}
Hsr = {sr, r^2sr} = {sr,sr^3}
so D4/H = {H,Hr,Hs,Hsr}.
now...make a multiplication table. is D4/H abelian? is it cyclic? what are the orders of each coset?
According to Lagrange's theorem every element must have an order that divides 4.
So, if we take any non-identity element, the order will be be 2 or 4.
If there is one of order 4, then the group is cyclic.
And all cyclic groups are abelian
all of this is true, but you have yet to convince me that you have calculated ANY of the orders of {H,Hr,Hs,Hsr}.
you seem to be convinced this group has an element of order 4. i ask you: which one?