00:01
Okay, so let's solve part a of our exercise.
00:05
Well, here we need to show that h is a normal subgroup of g.
00:12
How can we do this? well, we are gonna show, so we show that for every a belonging to gl2 of r and b belonging to h, we have a to the negative 1 multiplied by b multiplied by a belonging to h.
00:50
Well, this is pretty easy.
00:53
Indeed, we just need to compute the determinant of this matrix here.
00:58
I'm gonna call it x.
01:00
Now, what is the determinant of x? easy.
01:04
The determinant of x is the determinant of a to the negative 1, which is the determinant of a to the negative 1, multiplied by the determinant of b multiplied by the determinant of a.
01:17
So we can simplify and we get the determinant of b, which is equal to 7 to the k for some integer k.
01:31
Perfect.
01:32
So this shows that the determinant of x is 7 to the k, a power of 7, so x belongs to h.
01:43
Perfect.
01:44
Okay, now part b, part c of our exercise.
01:50
Let's solve c.
01:51
Well, here we need to show that if g is finite and non -abelian, finite and non -abelian, then g over the center is not cyclic.
02:23
Well, this is pretty easy...