Quotient group?

Question

Let G=Z4 x Z2 be a group, and H= < (0,1) > is a normal subgroup of G generated by (0, 1).

List the element of the quotient group G/H.


As i know that
G=Z4 x Z2
={ (0,0), (0,1), (1,0), (1,1), (2,0), (2,1), (3,0), (3,1)}

Then i don't have idea to find the element of quotient group G/H. Could someone help me how to find G/H.
Thanks.

Answer 1

H = { (0,1), (0,0) } ,which is isomorphic to Z2. Hence,G/H is isomorphic to Z4.

Answer 2

Okay, first of all let's remember the definition of the quotient group. The quotient group consists of all cosets of H in G, with the group operation (aH)(bH) = (ab)H.



In this example, first we have to find H. Well, the element (0,1) has order 2, and so H will have two elements (the other one is the identity), in other words H = {(0,0), (0,1)}.



The definition above, coupled with the enumeration of G which you give, shows that:



(Long form):

G/H = {(0,0)H, (0,1)H, (1,0)H, (1,1)H, (2,0)H, (2,1)H, (3,0)H, (3,1)H}



However some of these cosets are actually equal. To work this out, we use the following result: xH = yH if and only if x is an element of yH.



So, here goes:



(0,0)H = H = {(0,0), (0,1)} = (0,1)H (since (0,1) is an element of (0,0)H)

and similarly for the others:

(1,0)H = {(1,0)(0,0), (1,0)(0,1)} = {(1,0), (1,1)} = (1,1)H

(2,0)H = {(2,0)(0,0), (2,0)(0,1)} = {(2,0), (2,1)} = (2,1)H

(3,0)H = {(3,0)(0,0), (3,0)(0,1)} = {(3,0), (3,1)} = (3,1)H



Obviously if we had a much larger group then we could prove the pattern shown above as a general theorem, but it is not really worth it here.



Now, when we list the elements of G/H, we only need to take a representative for each coset as the coset leader. For example, we could just take all of the elements at the left-hand sides of the equations above.



So in summary:



G/H = {(0,0)H, (1,0)H, (2,0)H, (3,0)H} with the operation (aH)(bH) = (ab)H.

Answer 3

no longer a impressive deal to this one...For a nonidentity ingredient y of the quotient group G/N, y isn't the identification so it has order > a million. y is an ingredient of G, so y^2 is in N, so y^2 maps to identification of G/N. So it has order <=2. 2 is the sole integer that satisfies the two those.