In Fraleigh's A first course in abstract algebra, for example problem,
Compute (Z4 * Z6) / <(0,2)>.
These are my process.
Let H=<(0,2)> = {(0,2), (0,4), (0,0)}
Then
(0,0)+H = H (1,0)+H = {(1,2),(1,4),(1,0)} (2,0)+H = {(2,2),(2,4),(2,0)} (3,0)+H = {(3,2),(3,4),(3,0)} (0,1)+H = {(0,3),(0,5),(0,1)} (0,3)+H = {(0,5),(0,1),(0,3)} (0,5)+H = {(0,1),(0,3),(0,5)}
So there cosets are (0,0) (1,0) (2,0) (3,0) (0,1)
So (Z4 * Z6) / <(0,2)> is isomorphic to Z4*Z2
But here's the statement is the books.
*Here the first factor of Z4 of Z4*Z6 is left alone. The Z6 factor, on the other hand is essentially collapsed by a subgroup of order 3, giving a factor group in the second factor of order 2 that must be isomorphic to Z2.*
I can’t understand this statement that what ‘collapsed’ mean in this statement though I read previous part. Also what ‘left alone’ means.
And are there any easy way to understand 'coset', 'factor group(quotient group)' ?
