# When does a finite group $G$ have an element of order equal to its exponent?

The following question came up to my mind while I was playing with the exponent. There is no other motivation except plain curiosity:

Question. If $$G$$ is a finite group, denote by $$e(G)$$ its exponent. Can we characterize finite groups $$G$$ which have an element of order $$e(G)$$ ?

Examples and non examples.

• An abelian group $$G$$ has an element of order $$e(G)$$

• A $$p$$-group has an element of order $$e(G)$$

• More generally (as pointed out by @Brauer Suzuki), a finite nilpotent group $$G$$ has an element of order $$e(G)$$

• A non cyclic group $$G$$ whose Sylow subgroups are cyclic (eg $$D_n$$ for $$n$$ odd) has no element of order $$e(G)$$. Indeed, in this case, $$e(G)=\vert G\vert$$.

• For any $$e\geq 1$$ and any finite group $$H$$ such that $$e(H)\mid e,$$ $$G=H\times \mathbb{Z}/e\mathbb{Z}$$ has an element of order $$e(G)$$

• Finite groups $$G$$ satisfying the required property are not necessarily isomorphic to $$H\times \mathbb{Z}/e(G)\mathbb{Z}$$ for some suitable $$H$$, since the Heisenberg group of order $$p^3$$ is a counterexample.

Because of the last three items, I do not expect the answer to be easy (it is plausible that there is no satisfactory answer), but I am not an expert in group theory, so maybe there is a nice characterization in terms of linear representations, for example ? or in terms of derived subgroup ?

• Here $\exp$ is not the exponential, but rather the exponent of a group. Perhaps a different notation is useful, since the exponential map is often used for Lie algebras and Lie groups. This post shows your claim for abelian groups (as you said). Commented Nov 5, 2021 at 9:57
• Your first two examples can be generalized as "nilpotent groups $G$ have elements of order $\exp(G)$. Commented Nov 5, 2021 at 11:05
• @Dietrich Burde: I thought it was a standard notation. I changed it, following your advice. Commented Nov 5, 2021 at 12:30
• @Brauer Suzuki: indeed! I added this family of examples to my post. Commented Nov 5, 2021 at 12:30
• For any group $G$, $G^n$ has this property for large enough $n$. For example $A_5^3$ can be added to your list. Commented Nov 5, 2021 at 16:16

This is a partial answer to the question.

As remarked by @Brauer Suzuki, finite nilpotent groups $$G$$ have an element of order $$exp(G)$$. The same goes for all the sections of $$G$$: a section of $$G$$ is a group $$H/K$$, such that $$K \unlhd H \leq G$$. Now define $$\varphi(G)=\#\{g \in G: o(g)=exp(G)\}.$$ Since for a cyclic group $$G \cong C_n$$ of order $$n$$, this number equals $$\varphi(n)$$, the above number can be seen as a generalization of the Euler Totient function. Now, if $$G$$ is finite and nilpotent, then $$\varphi(G) \neq 0$$, and for all sections $$S$$ also $$\varphi(S) \neq 0$$. There is a remarkable theorem of $$M. T\check{a}rn\check{a}uceanu$$ (see here) that asserts that the converse is also true.

Theorem (M. Tarnauceanu, 2018) Let $$G$$ be a finite group. Then $$G$$ is nilpotent if and only if $$\varphi(S) \neq 0$$ for every section $$S$$ of $$G$$.

• I'm not sure what the introduction of the notation $\varphi$ accomplished here, but the result is nice! Commented Nov 5, 2021 at 13:00
• @Captain Lama: $\varphi$ of course counts the number of elements that have an order equal to the exponent of the group. As I explained above, for groups, it generalizes $\varphi$ (same notation) of Euler's Totient function. Commented Nov 5, 2021 at 13:09
• @DerekHolt: The exponents of all finite simple groups are known, and the orders of their elements too? Commented Nov 5, 2021 at 15:37
• This is a nice result, but it does not answer the question, since it is already known that nilpotent groups satisfy the desired property. However, would it be reasonable to think that if $G$ has an element of order $e(G)$, then it is isomorphic to a section of a nilpotent group ? Commented Nov 5, 2021 at 15:47
• @GreginGre No, because a section of a nilpotent group is nilpotent. Commented Nov 5, 2021 at 16:02

Here are some examples that put severe limits on what you can expect, in my opinion.

Example 1: Let $$G$$ be any finite group and let $$k$$ be the number of prime factors of $$|G|$$. Then the cartesian product $$G^k$$ has an element of order $$\exp(G) = \exp(G^k)$$. Indeed, let $$g_1, \dots, g_k \in G$$ be elements realising the maximal prime power orders in $$G$$. Then $$(g_1, \dots, g_k)$$ has order $$\exp(G)$$.

Note this generalizes the $$p$$-group example. The following example is similar.

Example 2: $$G = \operatorname{SL}_2(p)^3$$ for any prime $$p$$. Indeed, $$\operatorname{SL}_2(p)$$ has elements of order $$p-1$$, $$p$$, and $$p+1$$, so $$G$$ has an element of order $$\operatorname{lcm}(p-1, p, p+1) = \exp(G)$$.

Just one more source of (non-)examples: The groups with the desired property have a complete prime graph (not an "if and only if" though). I don't know want Dereck Holt deleted comments was, but I guess it was about simple groups. The prime graphs of simple groups are known, although I did not find an explicit claim about their connectedness.

• Derek commented that all nonabelian FSG are non-examples. Probably true, by inspecting the classification, but the Tarnauceanu result doesn't imply it. What is the prime graph of a group? Commented Nov 5, 2021 at 17:30
• The vertices of the prime graph are the prime divisors of $|G|$. Two distinct primes $p$ and $q$ are connected iff $G$ has an element of order $pq$. There is plenty of literature on that topic. Commented Nov 5, 2021 at 17:40
• I see. In the last line you mean "completeness" I think. Commented Nov 5, 2021 at 17:49
• No, my guess is that they are not even connected. Commented Nov 5, 2021 at 17:54
• I am confident that it could be proved that nonabelian finite simple groups and their close variations are non-examples by going through the simple groups. For example ${\rm PSL}(n,q)$ has no element of order $rs$, where $r$ and $s$ are Zsigmondy primes dividing $q^n-1$ and $q^{n-1}-1$. Commented Nov 6, 2021 at 8:23

Some small observations, too long for a comment.

Let $$G$$ be a finite group of order $$n$$ with prime factorization $$n = p_1^{k_1} \cdots p_t^{k_t}$$. Let $$P_i$$ be a Sylow $$p_i$$-subgroup of $$G$$.

Then $$\exp(G) = \exp(P_1) \cdots \exp(P_t)$$. The proof is an exercise. This was basically noted in one of the other answers, it follows then that the direct product $$G^t = G \times \cdots \times G$$ ($$t$$ times) contains an element of order $$\exp(G^t) = \exp(G)$$.

So the exponent is determined by the structure of the Sylow subgroups of $$G$$. Whether there exists an element of order $$\exp(G)$$ depends on the centralizers of elements of order $$\exp(P_i)$$.

For example, consider $$|G|= p_1^{k_1} p_2^{k_2}$$. Then $$G \times G$$ contains an element of order $$\exp(G \times G) = \exp(G)$$. And $$G$$ contains an element of order $$\exp(G)$$ if and only if for some $$x_1 \in P_1$$ of order $$\exp(P_1)$$, the centralizer $$C_G(x_1)$$ contains an element of order $$\exp(P_2)$$.