Question

Let G be a multiplicative group and H a finite subset G of nonempty, closed under multiplication. Show that H is a subgroup of G.

Answer 1

some one ca, help me ob this exercise

Answer 2

Multiplication in H is inherited from G, so that since associativity holds in G, it must also hold in H (that is, for all a,b,c in H, (a*b)*c = a*(b*c))
Next we must prove that the identity e is in H
H is finite, so it has some number (call it n) of elements; H is also nonempty, so it has an element a
Consider the mapping f from H into H defined by
f(b) = a*b
This mapping is injective, since
\[a*b_1 = a*b_2 \rightarrow b_1 = b_2\]which is evident from left multiplying by the inverse of a.
The mapping is also surjective; since it maps n elements injectively (that is, one-to-one) to n elements, each element of H must be mapped to.
Thus there exists some b in H such that f(b) = a
Then a*b = a, and b is the desired inverse e
Now let a be any element of H. By the previous argument, there must be some element b of H such that a*b = e, so that b is the desired inverse of a.
Since H is associative, contains the identity, and is closed under inverses, H must itself be a group, and so is a subgroup of G.

Answer 3

thank you ,the idea is clear.

Answer 4

But I thought of another demonstration I would like to show

Answer 5

Sure, go ahead!