Proving (Nb)^p = Nb^p for normal subgroup N and prime p

juaninf · 2011-03-24T12:20:58+00:00

Let G group and N subgroup normal from G if b \in{G} and p is prime number then (Nb)^p=Nb^p, Please help me with steps to this proof.

Level: Graduate
Thread starter juaninf
Start date Mar 24, 2011
Tags

Cauchy Groups Theorem

Join the discussion

Registration is free. Start your own thread to ask a follow-up.

1 reply · 2K views

juaninf

Thread starter
Mar 24, 2011

Let [tex]G[/tex] group and [tex]N[/tex] subgroup normal from [tex]G[/tex] if [tex]b \in{G}[/tex] and [tex]p[/tex] is prime number then [tex](Nb)^p=Nb^p[/tex],

Please help me with steps to this proof.

Physics news on Phys.org

Ultrathin materials could make quantum light circuits programmable
Spin rephasing helps quantum memories store single-photon states longer for future networks
Discovery confirms rare, switchable electrical property in widely used electronics material

Discussion

Char. Limit

Gold Member

Mar 24, 2011

You might find more help in the Homework Help forums. There are many more people there.

Proving (Nb)^p = Nb^p for normal subgroup N and prime p

Discussion

Similar threads

Graduate When do subgroups of larger order avoid containing a normal subgroup?

Graduate Proving 'If (2^n)-1 is Prime, Then n is Prime' Using Groups

If P is normal in H, and H normal in G, then P is normal in G

Graduate Is An a Normal Subgroup of Sn? A Proof and Explanation

Graduate Proving Normal Subgroup N is Central in Group G

Proof: There Exists a prime p such that p=< sqrt(n)

Graduate Why is the number of fixed points divisible by p in Cauchy's Theorem proof?

Graduate Sum of 1^k + 2^k + ... + (p-1)^k modulo a prime p

If a group G has order p^n, show any subgroup of order p^n-1 is normal

Graduate How can we show that x^m is in a normal subgroup N of G if |G/N| = m?