# Help using lagrange....

• Nov 10th 2009, 02:08 PM
ElieWiesel
Help using lagrange....
Hi I have this problem, and I cant seem to get it. It is not for a grade just for study practice. Let me know if you can do it. Thanks...

Let G be a group, |G|=\$\displaystyle p^{t}m \$ for a p a prime and (p,m) =1 and H \$\displaystyle \le\$ G with |H|=\$\displaystyle p^{t}\$. If K \$\displaystyle \le\$ G with |K|=\$\displaystyle p^{s}\$ and HK = KH, then show K\$\displaystyle \le\$ H.

Need to show K is a subgroup of H. Thanks again.
• Nov 10th 2009, 02:17 PM
Drexel28
Quote:

Originally Posted by ElieWiesel
Hi I have this problem, and I cant seem to get it. It is not for a grade just for study practice. Let me know if you can do it. Thanks...

Let G be a group, |G|=\$\displaystyle p^{t}m \$ for a p a prime and (p,m) =1 and H \$\displaystyle \le\$ G with |H|=\$\displaystyle p^{t}\$. If K \$\displaystyle \le\$ G with |K|=\$\displaystyle p^{s}\$ and HK = KH, then show K\$\displaystyle \le\$ H.

Need to show K is a subgroup of H. Thanks again.

What are you having trouble with exactly? What points aren't clear? Do you see why it is significant that \$\displaystyle (p,m)=1\$?
• Nov 10th 2009, 02:44 PM
ElieWiesel
Okay well maybe if I can explain what I have so far, maybe It will help.

HK= HK implies that HK \$\displaystyle \le\$ G

I also know that the |HK| = |H||K|/|H \$\displaystyle \cap\$ K|

I also know that |G| = |H| [G:H] which seems to imply that [G:H] = m by lagrange

Also I know that |HK| divides |G| by lagrange

Im thinking I have all the pieces put I dont know how to put them together right.