Abstract Algebra 9.2: Normal Subgroup Test

[@emmanuelogwuche1323]

thank you sir I am ur student here need more of ur teaching

[@johndavies729]

You stated at 5.58 that you showed that xHx^-1 is a subgroup of G. I can't find which video this is in. Please help

[@patrickjones1510]

It has been almost 3 years since I made the video, so I can't say I specifically remember. It is probably easier for me to type out the argument than it would be to go through and find it. Consider two elements of xHx^-1. We can consider them to be xax^-1 and xbx^-1 where a and b are elements of H. But then (xax^-1)*(xbx^-1) = xa(x^-1 * x)bx^-1 = xaebx^-1 = x(ab)x^-1. Since H is a subgroup, ab is in H, and so x(ab)x^-1 is in xHx^-1. Hence xHx^-1 is closed under the group operation. Similarly, since a is in H, so is a^-1, and then xa^-1x^-1 is in xHx^-1. But by the shoes-and-socks theorem, (xax^-1)^-1 = xa^-1x^-1, so xHx^-1 is closed under inverses. Since xHx^-1 is closed under the group operation and inverses, it must be a subgroup.

[@johndavies729]

@@patrickjones1510 Thanks! Could you please also explain the significance of G having "only one" subgoup? (As stated at 5.24)

[@patrickjones1510]

@@johndavies729 In general, (1) normal subgroups are incredibly important, and (2) checking via the definition can be very tedious. So any alternate method of determining whether a subgroup is normal is potentially a huge time-saver.