# Order, periods, and cosets

• March 15th 2011, 10:06 AM
Zalren
Order, periods, and cosets
Let $G$ be a group of order 36 and let $H$ be a normal subgroup of order 6. Let $y \in G$ be an element of period 4.
a) Explain why $y\not\in H$
Because the period of any element must divide the order of the gorup that the element is in. 4 | 36 so $y \in G$ but 4 does not divide 6 so $y \not\in H$

b) What is the order of the factor group $G/H$
$G/H$ order is 6 because order $H$ divides order $G$ = 6

c) Work out that $(yH)^4$ is trivial in $G/H$
$(yH)^4 = yHyHyHyH = yyyyHHHH = y^4H = eH$, which is the trivial coset

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

e) Prove that $y^2 \in H$
...

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

• March 15th 2011, 01:10 PM
tonio
Quote:

Originally Posted by Zalren
Let $G$ be a group of order 36 and let $H$ be a normal subgroup of order 6. Let $y \in G$ be an element of period 4.
a) Explain why $y\not\in H$
Because the period of any element must divide the order of the gorup that the element is in. 4 | 36 so $y \in G$ but 4 does not divide 6 so $y \not\in H$

b) What is the order of the factor group $G/H$
$G/H$ order is 6 because order $H$ divides order $G$ = 6

c) Work out that $(yH)^4$ is trivial in $G/H$
$(yH)^4 = yHyHyHyH = yyyyHHHH = y^4H = eH$, which is the trivial coset

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

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

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

e) Prove that $y^2 \in H$

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

Tonio

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