Exploration 2: Order of elements

Given that the order of $g$ is $n$, we know that $g^n$ is the lowest positive power of $g$ equal to the identity $e$. Then $g$ generates the subgroup $\<g\>=\{e,g,g^2,\ldots,g^{n-1}\}$, which has exactly $n$ elements, thus $|\<g\>|=n=o(g)$.

Recall that an element $g\in G$ can generate a subgroup $\<g\>\le G$. Since $\<g\>$ is a subgroup of a finite group, apply Lagrange’s Theorem to show that $|\<g\>|$ divides $|G|$. But $|\<g\>|=o(g)$, so $o(g)$ also divides $|G|$.

• Consider $G^p$, which is the direct product of $G$ with itself $p$ times. Its elements are $p$-tuples of elements from $G$. We will proceed by a counting argument, first by defining a set $S$ and providing two ways to count $|S|$.
• Let $S$ be the subset of $G^p$ where taking the product of all components in the $p$-tuple gives the identity $e_G$. That is, $$S=\{(g_1,g_2,\ldots,g_p)\in G^p\mid g_1g_2\ldots g_p=e_G\}$$
• To find $|S|$, note that the first $p-1$ components of each $p$-tuple in $G^p$ can be chosen arbitrarily from (finite) $G$, but the final component must be the inverse of the product of the first $p-1$ components. Therefore, $|S|=|G|^{p-1}$. Given that $|G|$ is divisible by a prime $p$, we have $|G|=pm$ for some $m$. Then $|S|=(pm)^{p-1}$.
• Any permutation of a $p$-tuple in $S$ is a $p$-tuple in $S$, since order doesn’t matter when we take product. Then consider two $p$-tuples equivalent when one is a cyclic permutation of the other. This divides $S$ into equivalence classes of two sizes: elements of the form $(g,g,\ldots,g)$ belong in an equivalence class of size $1$, while any other equivalence class must be of size $p$ because $p$ is prime.
• Towards contradiction, assume that $\{(e_G,e_G,\ldots,e_G)\}$ is the only equivalence class of size $1$. Then let $k$ be the number of equivalence classes of size $p$, so that $|S|=1+pk$.
• Therefore $|S|=(pm)^{p-1}=1+pk$. Taking mod $p$ gives $0\equiv 1\mod p$, contradiction.
• Therefore there must be more equivalence classes of size $1$, implying that $S$ contains some nontrivial $(g,g,\ldots,g)$. Membership in $S$ implies $g^p=e_G$, so $g\in G$ is an element of order $p$.

Proving this for $(g_1,g_2)\in G_1\times G_2$ is enough to prove the theorem by induction. Let $n=o(g_1)$ and $m=o(g_2)$, so that $g_1^n=e_1$ and $g_2^m=e_2$. Then $(g_1,g_2)^k=(g_1^k,g_2^k)=(e_1,e_2)$ implies $k$ is a multiple of both $n$ and $m$. Since order of an element is the least such multiple, the order of $(g_1,g_2)$ must be $\lcm(n,m)$.

• The intersection of two subgroups is a subgroup of both: $H\cap K\le H$ and $H\cap K\le K$.
• By Lagrange, $|H\cap K|$ is a divisor of both $|H|$ and $|K|$.
• Since $|H|$ and $|K|$ are coprime, their only shared divisor is $1$, so $|H\cap K|=1$, and therefore $H\cap K=\{e\}$.

There are two cases: either $|H|\ne |K|$ or $|H|=|K|=p$ for some prime $p$.

• If $H,K$ have different prime orders, then they have coprime orders and by the previous theorem the intersection is trivial.
• Otherwise, $|H|=|K|=p$ for some prime $p$. By Lagrange, $|H\cap K|$ is a divisor of both $|H|$ and $|K|$. Since $|H|=|K|=p$ is prime, $|H\cap K|$ is either $1$ or $p$. It can’t be $p$ since $H,K$ are distinct. So the intersection is trivial.

• The key is knowing that, given $g^n=e$, $$(hgh^{-1})^n=(hgh^{-1})(hgh^{-1})\ldots(hgh^{-1})=hg^nh^{-1}=heh^{-1}=e$$
• This shows that the order of $hgh^{-1}$ is at most the order of $g$.
• We can show that the order of $hgh^{-1}$ cannot be lower than the order of $g$ by contradiction.
• Suppose $m=o(hgh^{-1})<o(g)$, implying $(hgh^{-1})^m=e$. Then $$\begin{aligned} (hgh^{-1})^m&=e\\ hg^mh^{-1}&=e\\ g^m&=h^{-1}eh\\ g^m&=e\\ \end{aligned}$$ which is not possible since $m<o(g)$.
• Therefore the order of every $hgh^{-1}$ is exactly the order of $g$.

• Since $|G|$ is prime, $|G|\ge 2$ and therefore there is a non-identity element $g$ with order $\gt 1$.
• By Lagrange’s Theorem, the order of every element divides $|G|$. Since $|G|$ is prime and $o(g)\gt 1$, the order of $g$ can only be $|G|$.
• But that means $g$ generates $G$, therefore $G$ is cyclic.

• The center $Z(G)$ must be a subgroup, which by Lagrange’s Theorem must have order equal to one of $1,p,p^2$.
• Since $G$ is a $p$-group and therefore its center has order divisible by $p$, the center of $G$ cannot be order $1$.
• If the center is order $p$, then $|G/Z(G)|=p$ implies $G/Z(G)$ is cyclic. But that means $G$ is abelian.
• If the center is order $p^2$, it can only be the entire group $Z(G)=G$ which makes $G$ abelian by definition.

Recall that a group is abelian iff the quotient group $G/Z(G)$ is cyclic, and that a group is cyclic if it has prime order. Thus if $G/Z(G)$ has prime order, then $G$ is abelian.

Again by Lagrange’s Theorem, the order of the center must divide the order of the group $pq$, and thus the center can only be of order $1,p,q,pq$.

• If the center is order $pq$ then the group is abelian by definition.
• If the center is order $1$ then the group is centerless by definition.
• Otherwise, the center is prime $p$ or $q$, and therefore $G/Z(G)$ has prime order $pq/p=q$ or $pq/q=p$, which makes $G$ abelian.

• By Lagrange’s Theorem, $|G|=2p$ implies any element $g$ has order dividing $2p$, i.e. $o(g)\in\{1,2,p,2p\}$.
• If every non-identity element was order $2$ then $G$ is abelian, and the set $\{1,g,h,gh\}\iso K_4$ for any non-identity $g,h$ is closed and therefore a subgroup of $G$. By Lagrange, $|\{1,g,h,gh\}|=4$ must divide $|G|=2p$, a contradiction. So we must have some $g$ that is order $p$ or order $2p$.
  • If $g$ exists such that $o(g)=2p$, then $g$ generates $G$, so $G\iso C_{2p}$.
  • If $g$ exists such that $o(g)=p$, then $\<g\>$ is a subgroup of order $p$. WTS $G\iso D_p$.
    • $H=\<g\>$ is the unique subgroup of order $p$. To see this, consider another subgroup $K$ of order $p$. Since $H$ and $K$ are distinct prime order subgroups, their intersection is trivial. (proof) But then $|HK|=|H||K|/|H\cap K|=p^2$ (proof) which is greater than $|G|=2p$, contradiction.
    • Then for non-identity $k\in G\setminus \<g\>$, we know that $|\<k\>|\ne p$ since $\<g\>$ is the unique subgroup of order $p$. But since $k$ is not identity ($o(k)=1$), and $k$ does not generate $G$ ($o(k)=2p$), we must have $o(k)=2$ for every $k$ not in $\<g\>$.
    • Since $gk\notin \<g\>$, $o(gk)=2$ as well. This means both $k$ and $gk$ are self-inverse: $gk=(gk)^{-1}=k^{-1}g^{-1}=kg^{-1}$.
    • The group $G=\<g,k:g^p=k^2=e,gk=kg^{-1}\>$ is exactly (isomorphic to) the dihedral group $D_p$.

The order of the generator of $C_n$ must be $n$, so the exponent is at least $n$. By Lagrange’s theorem, the order of every element of $C_n$ divides its order $|C_n|=n$, so the exponent is at most $n$. Therefore the exponent is exactly $n$.

• Proving this for $G=G_1\times G_2$ is enough to prove the theorem by induction.
• If either of $G_1$ or $G_2$ have infinite exponent, then LCM of their exponents is infinite, which is equal to the exponent of $G$ (infinity), and we are done.
• Otherwise, assume $G_1$ or $G_2$ have finite exponents.
• For every $(g_1,g_2)$ of $G$, by definition of exponent, we know that $g_1^{\exp(G_1)}=e_{G_1}$ and $g_2^{\exp(G_2)}=e_{G_2}$.
• $\exp(G)$ is the least $k$ such that $(g_1,g_2)^k=(g_1^k,g_2^k)=(e_{G_1},e_{G_2})=e_G$. For the middle equality to hold, $k$ must be a multiple of both $\exp(G_1)$ and $\exp(G_2)$. Since $\exp(G)$ needs to be the least such $k$, we have $\exp(G)=k=\lcm(\exp(G_1),\exp(G_2))$.

This combines the previous two theorems. The exponent of a direct product is the LCM of the exponent of each $C_{k_i}$, which is $k_i$, thus the above statement follows.

If we take $g_i$ as the generator of each cyclic group, then $(g_1,g_2,\ldots,g_n)\in G$ has order equal to the LCM of the orders of each $g_i$, i.e. the LCM of all $k_i$, which is exactly the exponent of $G$.

Since $G$ is a direct product of cyclic groups each of order $k_i$, we have $|G|=\prod_{i=1}^n k_i$.

• ($\to$)
  • Since $G$ is cyclic, it has a generator $g=(g_1,g_2,\ldots,g_n)$ of order $|G|$, where each $g_i\in C_{k_i}$ generates $C_{k_i}$.
  • Since the order of a direct product element $(g_1,g_2,\ldots,g_n)$ must be the LCM of the order of each component $g_i$, we have $o(g)=|G|=\lcm(k_1,k_2,\ldots,k_n)$.
  • But then $|G|=\prod_{i=1}^n k_i$ (above) implies $\prod_{i=1}^n k_i=\lcm(k_1,k_2,\ldots,k_n)$. Since the LCM of each $k_i$ is equal to the product of each $k_i$, the $k_i$ must be pairwise coprime.
• ($\from$)
  • By the previous proof, there is an element $g\in G$ whose order is equal to the exponent of $G$, i.e. the LCM of each $k_i$. But since each $k_i$ are pairwise coprime, their LCM is equal to their product. That means $o(g)=\prod_{i=1}^n k_i=|G|$, which implies $g$ generates $G$, so $G$ is cyclic.