Let G = (T,*,e) be a group and U T. (a) Show that G\' = (U, *, e) is a subgroup

Question

Let G = (T,*,e) be a group and U T.

(a) Show that G' = (U, *, e) is a subgroup of G

(if and only if the following fact holds: for all x, y T, if x U and y U then x* y-1 U.)

(b) Assume that U is finite. Show that G' = (U, *, e) is a subgroup of G if and only if the set U is closed under the group operation *.

Hint for (a):

To show the other implication, it is sufficient to show the following three properties: (1) e U, and (2) if x U then x-1 U, and finally that (3) U is closed under the group operation *. Show them in this very order, by choosing x and y suitably.

Hint for (b):

As in the proof of Cayley’s theorem, for any x T, the mapping Px : T T defined as Px (y) = x * y is a bijection of T. Observe that if x U and U is closed on * then this function Px has the property that Px (U) U . If so then may be considered as a function Px : U U . Show that if U is finite and x U then this function Px : U U , is onto U. (It apparently is a bijection of U)

We need to show the equivalence but one implication follows immediately from the definition of a subgroup. To show the other implication, it is sufficient to show the three properties as for part 1).

The property of closeness is assumed to hold. To begin showing the remaining two properties, let x U.

Since function Px:U U is onto U, there exists z U such that Px (z) = x . What is this z?

This is Discrete Structure class.

Explanation / Answer

Let G = (T,*,e) be a group and U T.
Show- G' = (U, *, e) is a subgroup of G.

G' will be the subgroup of G when there will be two condition will met

1)Closure- If a,b belongs to U then ab belongs to U.
2)Inversee-for any a belongs to ,inverse of a belongs to U

Suppose G' is a sub group of G.
Then G will be the subset of G'.
So for associativity-
a,b,c belongs to G'              a,b,c belongs to G
     a(bc)=(ab)c
Identity-

a belongsto G'     Inverse of a belongs to G'
a(inverse(a)) belonfgs to G'     a(inverse(a))=e belongs to G'

Let G' be a subgroup of G
Since G' is itself a group,it must be closed under the operation *
Also it must be   true that G' contains the inverse of any of its elements.

So by Identity and associativity which is explained above proves that G' is a sbugroup of G.

2)
Assume that U is finite. Show that G' = (U, *, e) is a subgroup of G if and only if the set U is closed under the group operation *.

U is a finite subset of G.And we prove that G' is sub group of G.
1. We know that U is closed under the operation.
2.We know that U is finite.

So if x belongs to U then note square of(x),cube of(x).....so.on are all in U.
But H is finite ,there exist(m,n) belongs to (positive integers),m !=n such that X to the power of m=X to the power of n
So assume without loss of generality(wlog) that m>n.
Note X to the power(m-n)=e,so X to the power(m-n)=inverse of X.
Since m-n>0,so m-n>=1 (we can take)..

If m-n=1, Inverse of(X)=X to the power 0=e belongs to U.

If m-n>1, Inverse of(X)=X to the power (m-n)-1 belongs to U.
So U is colosed under inverse.
Therefore we can say that G' is a subgroup of G under the group operation.

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