Show that any subsemigroup of a finite group G is a subgroup

Show that any subsemigroup of a finite group G is a subgroup of G. Is this statement true for subsemigroups in any infinite group ?

Let (G, •) be a finite group and (H, •) be a subsemigroup of (G, •).

Therefore, H be a subset of G and a•(b•c) = (a•b)•c for all a, b, c H

By the property of semigroup, a H a^m H for all m N

N be the set of natural numbers.

a^m = a•a•…•a (m times)

Since G is finite and H be a subset of G, H is also finite.

Therefore, there must exists some natural number m and n (m > n), such that a^m = aⁿ for a H

Now, a H a G a^m, aⁿ G a^{m – n} G (as G is a group)

Also, a^m = aⁿ a^{m – n} = e (e be the identity element of G)

This implies, e H (because, a H a^m H for all m N)

Therefore, (H, •) contains the identity element.

We need to test whether the inverse exists, i.e. to prove that a H a^-1 H

If a = e, then a^-1 = a and we are done.

Let a e. Consider the sequence a, a², a³, …

Since, H is finite and close under the binary operation •, all positive powers of a are in H, not all of those elements are distinct.

Say, aⁱ = a^j and i > j

Then, a^{i – j} = e; and since a e, i – j > 1

Thus, a^{i – j} = a• a^{i – j – 1} = e

Therefore, a^{i – j – 1} = a^-1

But, i – j – 1 1 a^{i – j – 1} H a^-1 H

Therefore, H is a subgroup of G (Proved)

2^nd Part: The statement is not true because (Z, .) be a subsemigroup of the group (R*, .) but (Z, .) is not a subgroup of (R*, .)

Z be the set of all integers. R* be the set of all non-zero real numbers, i.e. R* = R – {0} and . represent usual multiplication.