Recent content by xmcestmoi

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 ?

I was just confused by a problem :( But I get your point.Thank you!

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?

Thank you :) I will try to come up with an isomorphism from <(3,2)> to Z9⊕Z4

Thank you :) I did not occur to me before

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?

Please help with this question: let ∣G∣=15. If G has only one subgroup of order 3 and only one of order 5, prove that G is cyclic. Thank you!