Let G be a group. and H is a finite subset of G which is closed under the binary operation of G.
Can anyone suggest a book?
Since $H$ is finite it follows that for any $a in H$ $a^{n}=a^{m}$ for some $n,m in N$ (powers of $a$ eventually form a cycle). Hence $x*a^{l}=x$ for some $x$ and $H$ contains the inverse of any $a$ and the identity.
Every element $x$ of $H$ induces by left multiplication an injective (thus bijective as $H$ is finite) map of $H$ into itself. As $x$ lies in $H$, one can find an element $y$ of $H$ such that $xy=x$, so $y=1$, and thus $H$ contains the identity element. Similarly, one can find $z$ in $H$ such that $xz=1$, thus $z=x^{-1}$ lies in $H$.
Sorry, I tried to ask, related questions can we find somewhere?
i suggest a contemporary abstract algebra for GALLIAN ( ANY edition)
I just thought I should offer some incisive comment on the (correct) proof provided by Yassine Guerboussa. So let H be a finite subset of a group G. The hypothesis on H says that H is a finite semi-group. However, an elementary result in group theory says that a finite semi-group H is a group if and only if H satisfies cancellation laws. But H , as a subset of a group G satisfies cancellation laws since G does. Hence H is a group.
Let H be finite subset of a group G which is closed under multiplication. Let h be an element of H .Let S ={ h^n s.t . ,n is an integer } .Clearly the elements of S are not distinct as S is a subset of a finite set. So there are positive integers n ,ms.t.
h^n= h^m with n> m or n< m say n > m .Now h^(n- m) =1 the multiplicative identity of G.Then h^(n -m-1) h =1 . Thus h in H has a multiplicative inverse in h as h^(n-m-1) is in H knowing that H is closed under multiplication by assumption. So H is closed under inverses and under multiplication .hence H is a sub group of G.
@Jaber Abujokha : It is much simpler, any element $h$ of $H$ has a finite order, say $h^n=1$, thus $h^{n-1}$ which is the inverse of $h$, lies in $H$ by the assumption that $H$ is closed under multiplication.
(one would be careful about $H$ is a subgroup whenever it is closed under multiplication and inverses, as this is false if $H$ is empty)
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