Let $\displaystyle G$ be a group of order 36 and let $\displaystyle H$ be a normal subgroup of order 6. Let $\displaystyle y \in G$ be an element of period 4.

a) Explain why $\displaystyle y\not\in H$

Because the period of any element must divide the order of the gorup that the element is in. 4 | 36 so $\displaystyle y \in G$ but 4 does not divide 6 so $\displaystyle y \not\in H$

b) What is the order of the factor group $\displaystyle G/H$

$\displaystyle G/H$ order is 6 because order $\displaystyle H$ divides order $\displaystyle G$ = 6

c) Work out that $\displaystyle (yH)^4$ is trivial in $\displaystyle G/H$

$\displaystyle (yH)^4 = yHyHyHyH = yyyyHHHH = y^4H = eH$, which is the trivial coset

d) Prove $\displaystyle yH$ has period 2 in $\displaystyle G/H$

Neither $\displaystyle y\in H$ (why?) nor $\displaystyle ord(yH)=1$ (why?These two claims are, in

fact, one and the same thing...), but still $\displaystyle (yH)^2=1\in G/H$, thus...
e) Prove that $\displaystyle y^2 \in H$

This follows at once from (c) and from $\displaystyle xH=H\Longleftrightarrow x\in H$

Tonio
For some reason when it comes to doing periods of cosets I get confused. Can anyone help me out?

Thanks in advance,