This is just something I conjectured after playing around with some examples of groups.
Prove that a subgroup of a cyclic group is cyclic.
This is just something I conjectured after playing around with some examples of groups.
Prove that a subgroup of a cyclic group is cyclic.
Let $G=\langle a \rangle$ be cyclic group generated by 'a'& H be a subgroup of group G let m be the smallest positive integer such that $a^m \in H$ then clearly $\langle (a^m)^q \rangle\ \in H$ $\langle a^m\rangle\subset H$
let $a^t\in H$
applying division algorithm for integer t&m $\exists q,r\in Z$ such that
$t=m.q+r$, where $0\leq r
$\implies r =t-mq$
$\implies a^r=a^t.a^{-mq}$
$\implies a^t.(a^m)^{-q}\in H$
$\implies a^r \in H$ , r< m
If $r\neq 0$ then r is the smallest positive integer such that $a^r\in H$
Which is a contradiction {$\because$ m is the smallest positive integer }
\therefore $r=0 $
$\implies t=m.q $
$\implies a^t=a^m.q$
$\implies a^m.q \in \langle a^m \rangle$
$\implies H \subset \langle a^m \rangle$
$\therefore H=\langle a^m \rangle$
Thus, Subgroup of cyclic group is cyclic
Hint. Let $G=\langle g \rangle$ be a cyclic group. Let $H$ be a subgroup of $G$. If $H$ is trivial, then $H=\langle e\rangle$. Otherwise, prove that $H=\langle g^k\rangle$, where $k$ is the smallest positive integer such that $g^k\in H$.
You'll have to prove that if $H\neq\{e\}$, then there is such a $k$, and that the subgroup it generates is equal to $H$.