Show that
(1) there is no homomorphism from A4 onto a group of order 2, 4, or 6
(2) there is a homomorphism from A4 onto a group of order 3.
Can anyone please give me some hints ?
another question concerning a group of order pq
if the order of a group G is pq, where p,q are distinct prime numbers,
then G has two normal subgroups, one has order p, and the other has order q.
Is this statement true?
I am trying to do the followin 2 problems but not sure if I am doing them correct.
Anyone please have a look...
1. In Z40⊕Z30, find two subgroups of order 12.
since 12 is the least common multiple of 4 and 3, and 12 is also least common multiple of 4 and 6.
take 10 in Z40, and 10...
isnt it true that any element in G, say an element x will generate a cyclic group with order = the order of x ?
and then going back to the original question , which says "only one subgroup of order 3"
so <x> is "the" subgroup H
I suspect that
if x has order 3, then it generates H
and similarly, if x has order 5, then it generates K
in either case, x will be in H or K
contracting previous fact that x of G is in neither H nor K
Thank you!
1. for a group G of order pq to be cyclic, it needs to be generated by an element with order pq.
2. If H is the only subgroup of order p and K is the only subgroup of order q , then the size of the set HUK is P+q-1, because the identity element of G is the only element in both H...
I know that any group with a prime order is cyclic.
But what about a group with an order pq, where p and q are distinct prime numbers?
Is this group also cyclic?