Exponent of Group Hey! :o - Answers to Questions

In summary, the conversation discusses the concept of exponent in group theory. It is defined as the smallest positive integer such that each element of the group raised to this power yields the identity element. It is shown that the exponent divides the order of the group, and in the case of a cyclic group, the exponent is equal to the order. The conversation then moves on to discussing the exponent of the cartesian product of groups, which is shown to be a multiple of the least common multiple of the exponents of the individual groups.
  • #1
mathmari
Gold Member
MHB
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Hey! :eek:

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)
 
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  • #2
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.
 
  • #3
Euge said:
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)
Euge said:
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)
Euge said:
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)
 
  • #4
mathmari said:
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)
mathmari said:
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 $r$.
 
  • #5
Euge said:
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)
 
  • #6
mathmari said:
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.
 
  • #7
Thank you very much! (Happy)
 
  • #8
Euge said:
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)
 
  • #9
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, $m = \exp(G)$.
 
  • #10
Euge said:
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)
 
  • #11
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 $m$.
 
  • #12
Euge said:
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)
 
  • #13
I got stuck right now... Which is the difference between the order of a group and the exponent of a group? (Wondering)
 
  • #14
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 $2$.
 
  • #15
And why does it hold that $g^{|G|}=e$, for $g\in G$ ? (Wondering)
 
  • #16
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 $k$, $g^{\lvert G\rvert} = g^{km} = (g^m)^k = e^k = e$.
 
  • #17
Euge said:
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)
 

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

What is the exponent of Group Hey!?

The exponent of Group Hey! is the number or symbol placed above and to the right of the group's name, indicating the number of times the group is being multiplied by itself.

How is the exponent of Group Hey! calculated?

The exponent of Group Hey! is calculated by counting the number of members in the group and raising it to the power of the group's rank or position in the hierarchy.

What is the significance of the exponent in Group Hey!?

The exponent in Group Hey! represents the growth and expansion of the group as it gains more members and influence. It also highlights the importance and strength of the group within its community.

Can the exponent of Group Hey! change?

Yes, the exponent of Group Hey! can change if the group gains or loses members, or if its rank or position in the hierarchy changes. It is a dynamic representation of the group's current size and status.

Is the exponent of Group Hey! unique to this group?

Yes, the exponent of Group Hey! is unique to this group and cannot be applied to any other group. Each group has its own unique exponent, reflecting its individual growth and influence.

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