MHB Exponent of Group Hey! :o - Answers to Questions

[mathmari]

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)

 

[Euge]

Hi mathmari,

Your proof looks good, but it can be shortened as follows:

Let $|G| = q \cdot \exp(G) + r$, where $q$ and $r$ are integers with $0 \le r < \exp(G)$. For every $g\in G$, $g^r = g^{|G|}(g^{\exp(G)})^{-q} = e$. By minimality of $\exp(G)$, $r = 0$; therefore, $|G| = q\cdot \exp(G)$, proving $\exp(G)\, |\, |G|$. If $G$ is cyclic, let $x$ be a generator of $G$. Since $x^{\exp(G)} = 1$, the order of $x$ divides $\exp(G)$; as the order of $x$ is $|G|$, then $|G|\, |\, \exp(G)$. It was already shown that $\exp(G)\, |\, |G|$, so we have $\exp(G) = |G|$.

To answer your last question, let $m = \operatorname{lcm}\{\exp(H_1),\ldots, \exp(H_n)\}$. Set $\exp(G) = mq + r$, for some integers $q$ and $r$ with $0 \le r < m$. Take an arbitrary element $(h_1,\ldots, h_n)\in G$. For $1 \le i \le n$, let $e_i$ be the identity of $H_i$. Then for all $i$, $h_i^m = (h_i^{\exp(H_i)})^{m/\exp(H_i)} = e_i^{m/\exp(H_i)} = e_i$. Consequently, for every $i$, $h_i^r = h_i^{\exp(G)}(h_i^m)^{-q} = e_i$. Therefore, $(h_1,\ldots, h_n)^r = (e_1,\ldots, e_n)$. Suppose $r > 0$. Since the element $(h_1,\ldots, h_n)$ was arbitrary, it follows that $\exp(G)$ divides $r$. In view of the equation $\exp(G) = mq + r$, we deduce $m$ divides $r$, contradicting the inequality $r < m$. Therefore, $r = 0$ and $m$ divides $\exp(G)$, as desired.

 

[mathmari]

Your proof looks good, but it can be shortened as follows:

Let $|G| = q \cdot \exp(G) + r$, where $q$ and $r$ are integers with $0 \le r < \exp(G)$. For every $g\in G$, $g^r = g^{|G|}(g^{\exp(G)})^{-q} = e$. By minimality of $\exp(G)$, $r = 0$; therefore, $|G| = q\cdot \exp(G)$, proving $\exp(G)\, |\, |G|$. If $G$ is cyclic, let $x$ be a generator of $G$. Since $x^{\exp(G)} = 1$, the order of $x$ divides $\exp(G)$; as the order of $x$ is $|G|$, then $|G|\, |\, \exp(G)$. It was already shown that $\exp(G)\, |\, |G|$, so we have $\exp(G) = |G|$.

Ah ok! (Smile)

To answer your last question, let $m = \operatorname{lcm}\{\exp(H_1),\ldots, \exp(H_n)\}$. Set $\exp(G) = mq + r$, for some integers $q$ and $r$ with $0 \le r < m$. Take an arbitrary element $(h_1,\ldots, h_n)\in G$. For $1 \le i \le n$, let $e_i$ be the identity of $H_i$. Then for all $i$, $h_i^m = (h_i^{\exp(H_i)})^{m/\exp(H_i)} = e_i^{m/\exp(H_i)} = e_i$. Consequently, for every $i$, $h_i^r = h_i^{\exp(G)}(h_i^m)^{-q} = e_i$.

We have that $(h_1,\ldots, h_n)\in G$. From that it follows that $ h_i^{\exp(G)}=e$, right? (Wondering)

Therefore, $(h_1,\ldots, h_n)^r = (e_1,\ldots, e_n)$. Suppose $r > 0$. Since the element $(h_1,\ldots, h_n)$ was arbitrary, it follows that $\exp(G)$ divides $r$. In view of the equation $\exp(G) = mq + r$, we deduce $m$ divides $r$, contradicting the inequality $r < m$. Therefore, $r = 0$ and $m$ divides $\exp(G)$, as desired.

Since $\exp(G)$ divides $r$ and $\exp(G) = mq + r$ how do we conlcude that $m$ divides $r$ ? Can it not be that $q$ divides $r$ ? (Wondering)

 

[Euge]

We have that $(h_1,\ldots, h_n)\in G$. From that it follows that $ h_i^{\exp(G)}=e$, right? (Wondering)

If by $e$ you mean $e_i$, then yes that's right. (Yes)

Since $\exp(G)$ divides $r$ and $\exp(G) = mq + r$ how do we conlcude that $m$ divides $r$ ? Can it not be that $q$ divides $r$ ? (Wondering)

If $\exp(G)$ divides $r$, then since $\exp(G) = mq + r$, then $mq + r$ divides $r$. Hence, $mq$ divides $r$. Since $m$ divides $mq$ and $mq$ divides $r$, then $m$ divides $r$.

 

[mathmari]

If by $e$ you mean $e_i$, then yes that's right. (Yes)


If $\exp(G)$ divides $r$, then since $\exp(G) = mq + r$, then $mq + r$ divides $r$. Hence, $mq$ divides $r$. Since $m$ divides $mq$ and $mq$ divides $r$, then $m$ divides $r$.

I got it! (Happy) If the groups $H_1, \ldots , H_n$ are cyclic we have from above that $\exp (H_i)=|H_i|$.
So, we have that $\text{lcm}\{|H_1|,\ldots, |H_n|\}$ divides $\exp (G)$, right? (Wondering)

 

[Euge]

If the groups $H_1, \ldots , H_n$ are cyclic we have from above that $\exp (H_i)=|H_i|$.
So, we have that $\text{lcm}\{|H_1|,\ldots, |H_n|\}$ divides $\exp (G)$, right? (Wondering)

Yes, that's correct.

 

[mathmari]

Thank you very much! (Happy)

 

[mathmari]

If $\exp(G)$ divides $r$, then since $\exp(G) = mq + r$, then $mq + r$ divides $r$. Hence, $mq$ divides $r$. Since $m$ divides $mq$ and $mq$ divides $r$, then $m$ divides $r$.

I thought about this again and now I got stuck. Knowing that $mq + r$ divides $r$ how do we get that $mq$ divides $r$ ? (Wondering)

 

[Euge]

That was an error on my part. It should go the other way: if $r$ divides $mq + r$, then $r$ divides $mq$. I will alter the proof the second theorem as follows (using the same meaning of $m$):

For every $(h_1,\ldots, h_n)\in G$, $(h_1,\ldots, h_n)^m = (h_1^m,\ldots, h_n^m) = (e_1^{m/\exp(H_1)},\ldots, e_n^{m/\exp(H_n)}) = (e_1,\dots, e_n)$. Furthermore, if $k$ is a positive integer such that $(h_1,\ldots, h_n)^k = (e_1,\ldots, e_n)$ for all $(h_1,\ldots, h_n)\in G$, then for each $i\in \{1,\ldots, n\}$, $h_i^k = e_i$ for all $h_i\in H_i$. Hence, $\exp(H_i)$ divides $k$ for all $i$. This means $m$ divides $k$, so $m \le k$. This shows that $m$ is the least positive integer for which $(h_1,\ldots, h_n)^m = (e_1,\ldots, e_n)$ for all $(h_1,\ldots, h_n) \in G$. Thus, $m = \exp(G)$.

 

[mathmari]

Hence, $\exp(H_i)$ divides $k$ for all $i$. This means $m$ divides $k$, so $m \le k$.

How do we get that $m$ divides $k$ ? (Wondering)

We have that $\exp(H_i)$ divides $k$ and $\exp(H_i)$ divides $m$, or not? (Wondering)

 

[Euge]

I have have shown that $k$ is a common multiple of the $\exp(H_i)$. Since $m$ is the least common multiple of the $\exp(H_i)$, then $k$ is a multiple of $m$.

 

[mathmari]

I have have shown that $k$ is a common multiple of the $\exp(H_i)$. Since $m$ is the least common multiple of the $\exp(H_i)$, then $k$ is a multiple of $m$.

Ahh I see! Thank you so much! (Smile)

 

[mathmari]

I got stuck right now... Which is the difference between the order of a group and the exponent of a group? (Wondering)

 

[Euge]

The order of a group $G$ is equal to the number of elements in $G$, if $G$ is finite. If $G$ is infinite, we say that $G$ has infinite order.

The exponent of $G$ is defined as the least positive integer $n$ such that $g^n = e$ for every $g \in G$.

To see the difference, consider the Vierergruppe $\Bbb Z/2\Bbb Z \times \Bbb Z/2\Bbb Z$. Its order is $4$ (as the name suggests), but its exponent is $2$, since every non-identity element has order $2$.

 

[mathmari]

And why does it hold that $g^{|G|}=e$, for $g\in G$ ? (Wondering)

 

[Euge]

That's when $G$ is finite, and it follows from Lagrange's theorem: If $H$ is a subgroup of a finite group $G$, then $\lvert H\rvert$ divides $\lvert G\rvert$. Indeed, if $G$ is finite and $g\in G$ has order $m$, then the cyclic subgroup generated by $g$ has order $m$, and hence $m$ divides $\lvert G\rvert$ by Lagrange. Letting $\lvert G\rvert = km$ for some integer $k$, $g^{\lvert G\rvert} = g^{km} = (g^m)^k = e^k = e$.

 

[mathmari]

That's when $G$ is finite, and it follows from Lagrange's theorem: If $H$ is a subgroup of a finite group $G$, then $\lvert H\rvert$ divides $\lvert G\rvert$. Indeed, if $G$ is finite and $g\in G$ has order $m$, then the cyclic subgroup generated by $g$ has order $m$, and hence $m$ divides $\lvert G\rvert$ by Lagrange. Letting $\lvert G\rvert = km$ for some integer $k$, $g^{\lvert G\rvert} = g^{km} = (g^m)^k = e^k = e$.

I understand! Thank you so much! (Smile)