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Let $C_m$ be a finite cyclic group of order $m\ge 1$. Therefore $|C_m|=m$ and there is an element $g\in C_m$ such that, in multiplicative notation, we have:

$$C_m=\{g^n\mid 0\le n\le m-1\}$$

We say that $g$ generates $C_m$ and we write $C_m=\langle g\rangle$.

Let $G$ be a group with identity element $e$. An element $h\in G$ has order $k\ge 1$, denoted $|h|=k$, if $|\langle h\rangle|=k$. Therefore $k$ is the smallest positive integer with $h^k=e$.

An abelian group $G$, where we write the composition law using the multiplicative notation, is one such that $ab=ba$ for every $a,b\in G$. Notice that a finite cyclic group $C_m$ is abelian.

Which of the following is ALWAYS true?


If $G$ is a finite abelian group and $\langle g\rangle$ is the largest cyclic subgroup of $G$, then the order of any other cyclic subgroup of $G$ divides $|\langle g\rangle|=|g|$.


If $G$ is an abelian group, then for every $g$, $h\in G$, with $g$, $h$ of finite order and $gh\not =e$, we have $|gh|=|g||h|$.


If $g\in G$ has finite order $k$, then $g^\ell$ has order $k$ for every $\ell$.


The abelian group ${\mathbb R}^\ast$ of non-zero real numbers under multiplication is cyclic.


The unit circle $S^1=\{t\in {\mathbb C}\mid |t|=1\}$ under multiplication is cyclic.

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