3 Let G be a group and let p be a prime number such that pg

3. Let G be a group and let p be a prime number such that pg 0 for every element g E G. a. If G is commutative under multiplication, show that the mapping is a homomorphism b. If G is an Abelian group under addition, show that the mapping f(x) x is a homomorphism

Solution

Lat G be group and p be a prime number such that for every element g of G pg=0

(a)

let G is commutative group under multiplication

i,e for x,y in G

x.y=y.x

define a function f:G---->G such that

   f(x)=x^p

to show that f is homomorphism

i.e to show 1) f(x+y)=f(x)+f(y)

                 2) f(xy)=f(x)f(y)   for x ,y in G

for 1)

let x,y in G

and p is prime

consider

f(x+y)=(x+y)^p                 (by defination of f )

         =x^p+px^p-1y+(p(p-1)/2!)x^p-2y²+(p(p-1)(p-2)/3!)x^p-3y³+--------------+y^p               (Binomial expantion)

         =x^p+y^p           because G is commutaive under multiplication & for prime p   pg=0 for all g in G

now 2)

cinsuder f(xy)=(xy)^p                    by defination of f

                    =x^py^p

                    = f(x)f(y)

therfor e f is homomorphism

b)

G is abilian group under addition

i.e a+b=b+a for a,b in G

define a function f:G---->G such that

   f(x)=x^p

to show that f is homomorphism

i.e to show 1) f(x+y)=f(x)+f(y)

                 2) f(xy)=f(x)f(y)   for x ,y in G

for 1)

let x,y in G

and p is prime

consider

f(x+y)=(x+y)^p                 (by defination of f )

         =x^p+px^p-1y+(p(p-1)/2!)x^p-2y²+(p(p-1)(p-2)/3!)x^p-3y³+--------------+y^p               (Binomial expantion)

         =x^p+y^p + px^p-1y+(p(p-1)/2!)x^p-2y²+(p(p-1)(p-2)/3!)x^p-3y³+------------- since isG abilian under addition

       = x^p+y^p+p[x^p-1y+((p-1)/2!)x^p-2y²+((p-1)(p-2)/3!)x^p-3y³+--------------+]   p is comman

       = x^p+y^p+pg                where g=x^p-1y+((p-1)/2!)x^p-2y²+((p-1)(p-2)/3!)x^p-3y³+--------------+    since g is group

       =x^p+y^p+0                     since pg=-0 g in G

       =x^p+y^p

        =f(x)+f(y)

f(x+y)=f(x)+f(y)

now 2)

cinsuder f(xy)=(xy)^p                    by defination of f

                    =x^py^p

                    = f(x)f(y)

therfor e f is homomorphism