# Factor Groups: What am I thinking about wrongly here?

1. Oct 26, 2014

### PsychonautQQ

say K is normal in G hence we have a factor group G/K.

let g be an element of G where |g| = n.

so Kg^n = K since g^n = 1.
and using the properties of factor groups, we know Kg^n = (Kg)^n
hence (Kg)^n = K
So we know that the order of Kg divides n.

Is this correct thinking? Factor groups are trippin me out

2. Oct 26, 2014

### Boorglar

Yes, this looks correct. In fact, this shouldn't be surprising, since by Lagrange's theorem, the order of the factor group G/K is |G|/|K| which divides |G| = n.
Thus the order of any element of G/K must divide |G/K| which divides n. You can easily show that a divisor of a divisor of n is also a divisor of n.

3. Oct 26, 2014

### PsychonautQQ

you said |G| = n. I originally said it was an element of G, |g| = n. Doesn't |G| = |g| iff G = <g> (G is a cyclic group generated by g?) The question doesn't say G is cyclic.

4. Oct 26, 2014

### Boorglar

Oh my bad I misread that the order of G was n. In that case the statement that |Kg| divides n does not directly follow from what I said, but it is still true by what you showed.