Abstract Algebra | Internal direct product of subgroups.

[@paul21353]

At 3:12 the order of the two h's on the LHS is wrong. It should be (h')^-1.h This does not change the argument.

[@algebraentodaspartes]

I love this videos. You are very inspiring professor. Thank you so much.

[@abnerandreymartinezzamudio3366]

Thank you for your hard work

[@azziahmed4721]

Thanks

[@maurocruz1824]

I got a question. Can I express the quaternion group as a internal direct product of subgroups?

[@Eye-vp5de]

I know I'm a year late, but: We know i,j,k anticommute with each other, so all of them should be in the same subgroup (otherwise hk=kh doesn't hold). But if i,j,k are in the same subgroup, so is -1 (i²), -i, -j, -k, 1 (as an identity element). So if Q8 is expressed as an internal direct product of two subgroups, than one of these subgroups is the whole group, so we're left with only one solution (product) - Q8{1} (also {1}Q8, but I would consider it as the same product, as internal direct product commutes). It's quite easy to see that any group can be expressed as an internal direct product of itself and a trivial subgroup.

[@omerbaba5754]

Let G=S_3 and let H is group of order 2 such that generator is 2 cycle and K is group of order 3 s.t generator is 3 cycle then this theorem does not satisfy so it is not true