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mathmari
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Hey!
For each group $G$, $\text{exp}(G)$ is the exponent of the group $G$, i.e., the smallest positive integer $k$, such that $g^k=e$ for each $g\in G$.
Let $G$ be a finite group.
I have shown that $\text{exp}(G)$ divides $|G|$, and if $G$ is cyclic, then $\text{exp}(G)=|G|$, as follows: We asssume that $\text{exp}(G)$ doesn't divide $|G|$, then $|G|=q\cdot \text{exp}(G)+r$ with $0<r<\text{exp}(G)$.
Let $g\in G$. We have that \begin{equation*}g^{|G|}=g^{q\cdot \text{exp}(G)+r}=\left(g^{\text{exp}(G)}\right )^qg^r \Rightarrow e=e^qg^r \Rightarrow g^r=e\end{equation*}
$r$ must be a multiple of the order of $g$, $\text{ord} (g)$.
Indeed, let $r=p\cdot \text{ord} (g)+c$ with $0<c<\text{ord} (g)$, then
\begin{equation*}g^r=\left (g^{\text{ord} (g)}\right )^pg^c\Rightarrow g^c=e\end{equation*}
A contradiction, since the order of $g$ is the smallest integer such that $g^{\text{ord} (g)}=e$, but $c<\text{ord} (g)$. Since $g$ is arbitrary, we have that $r$ is a multiple of $\text{ord} (g), \ \forall g\in G$.
But $\text{exp}(G)$ is the least common multiple of $\text{ord} (g), \ \forall g\in G$, i.e.e, it must hold that $\text{exp}(G)<r$. But we have that $r<\text{exp}(G)$. A contradiction.
So, it must be $r=0$. So, $\text{exp}(G)$ divides $|G|$.
Since $G$ is cyclic there is some $g\in G$ that generates $G$. Each element of $G$ can be written as $g^n$ for some $n\in \mathbb{Z}$ and for all these elements it holds \begin{equation*}\left (g^n\right)^{|G|}=\left ( g^{|G|}\right )^n=e^n=e\end{equation*} Since $\text{exp}(G)$ isthe smallest positive integer such that $g^{\text{exp}(G)}=e$, it follows that $\text{exp}(G)\leq |G|$.
The order of $g$ is $|G|$, i.e., $|G|$ is the smallest positive integer such that $g^{|G|}=e$ and since it holds that $g^{\text{exp}(G)}=e$, it must be $|G|\leq \text{exp}(G)$.
So, it follows that $\text{exp}(G)= |G|$.
Is everything correct? (Wondering) Now I want to show that if $G$ is the cartesian product of groups $\displaystyle{G=\prod_{i=1}^n H_i}$ then $\text{exp}(G)$ is a multiple of $\text{lcm} \{\text{exp}(H_1), \ldots , \text{exp}(H_n)\}$.
Could you give me a hint how we could show that? (Wondering)
For each group $G$, $\text{exp}(G)$ is the exponent of the group $G$, i.e., the smallest positive integer $k$, such that $g^k=e$ for each $g\in G$.
Let $G$ be a finite group.
I have shown that $\text{exp}(G)$ divides $|G|$, and if $G$ is cyclic, then $\text{exp}(G)=|G|$, as follows: We asssume that $\text{exp}(G)$ doesn't divide $|G|$, then $|G|=q\cdot \text{exp}(G)+r$ with $0<r<\text{exp}(G)$.
Let $g\in G$. We have that \begin{equation*}g^{|G|}=g^{q\cdot \text{exp}(G)+r}=\left(g^{\text{exp}(G)}\right )^qg^r \Rightarrow e=e^qg^r \Rightarrow g^r=e\end{equation*}
$r$ must be a multiple of the order of $g$, $\text{ord} (g)$.
Indeed, let $r=p\cdot \text{ord} (g)+c$ with $0<c<\text{ord} (g)$, then
\begin{equation*}g^r=\left (g^{\text{ord} (g)}\right )^pg^c\Rightarrow g^c=e\end{equation*}
A contradiction, since the order of $g$ is the smallest integer such that $g^{\text{ord} (g)}=e$, but $c<\text{ord} (g)$. Since $g$ is arbitrary, we have that $r$ is a multiple of $\text{ord} (g), \ \forall g\in G$.
But $\text{exp}(G)$ is the least common multiple of $\text{ord} (g), \ \forall g\in G$, i.e.e, it must hold that $\text{exp}(G)<r$. But we have that $r<\text{exp}(G)$. A contradiction.
So, it must be $r=0$. So, $\text{exp}(G)$ divides $|G|$.
Since $G$ is cyclic there is some $g\in G$ that generates $G$. Each element of $G$ can be written as $g^n$ for some $n\in \mathbb{Z}$ and for all these elements it holds \begin{equation*}\left (g^n\right)^{|G|}=\left ( g^{|G|}\right )^n=e^n=e\end{equation*} Since $\text{exp}(G)$ isthe smallest positive integer such that $g^{\text{exp}(G)}=e$, it follows that $\text{exp}(G)\leq |G|$.
The order of $g$ is $|G|$, i.e., $|G|$ is the smallest positive integer such that $g^{|G|}=e$ and since it holds that $g^{\text{exp}(G)}=e$, it must be $|G|\leq \text{exp}(G)$.
So, it follows that $\text{exp}(G)= |G|$.
Is everything correct? (Wondering) Now I want to show that if $G$ is the cartesian product of groups $\displaystyle{G=\prod_{i=1}^n H_i}$ then $\text{exp}(G)$ is a multiple of $\text{lcm} \{\text{exp}(H_1), \ldots , \text{exp}(H_n)\}$.
Could you give me a hint how we could show that? (Wondering)