1 Introduction
Graphs arising from groups have become an important topic in the last fifteen years (see for example [Reference Freedman4–Reference Lucchini and Nemmi6, Reference Rekvényi8]). In this note, we give a simplified proof of the following results.
Theorem 1.1. Let G be a finite group. Then, outside of its isolated vertices, the complement of the enhanced power graph of G is connected and its diameter is at most
$3$
.
Corollary 1.2. Let G be a finite group. Then, outside of its isolated vertices, the complement of the power graph of G is connected and its diameter is at most
$3$
.
The problem of the connectedness of the complement of the enhanced power graph was posed in [Reference Cameron2, Question 20]. We are aware of two distinct solutions of this problem: [Reference Abdollahi and Mohammadi Hassanabadi1, Proposition 3.2] and [Reference Ma, Doostabadi and Wang7, Theorem 1.6], both also showing that the unique connected component of such a graph has diameter at most
$3$
. Our proof identifies the largest independent set of the complement of an enhanced power graph, and exploits its properties to show connectedness and bounded diameter. We would also like to point out that [Reference Abdollahi and Mohammadi Hassanabadi1] predates [Reference Cameron2], but their contribution has remained unnoticed because their name for the complement of the enhanced power graph is different.
The connectedness of the complement of the power graph is proved in [Reference Cameron2, Theorem 9.9], while [Reference Cameron2, Question 19] asks whether or not its diameter is bounded, and an answer in the affirmative has been given in [Reference Ma, Doostabadi and Wang7, Theorem 1.5]. The novelty of our proof of Corollary 1.2 is that it is an easy consequence of Theorem 1.1.
We conclude this introduction by pointing out that both bounds in Theorem 1.1 and Corollary 1.2 are sharp. Indeed, for two distinct primes p and r, the complements of the power graph and of the enhanced power graph of
$C_{pr} \times C_{p^2}$
have diameter
$3$
(see [Reference Ma, Doostabadi and Wang7, Lemmas 2.6 and 3.3]).
2 Proof of Theorem 1.1
Let G be a finite group. The enhanced power graph of G is defined as the (undirected) graph
$\Gamma _{\mathrm {EP}}(G)$
of vertex set G, where two vertices are declared adjacent if they are contained in a common cyclic subgroup of G. We are interested in its complement, which we denote by
$\mathcal {C}\Gamma _{\mathrm {EP}}(G)$
.
Proof of Theorem 1.1.
Let G be a finite group and let
$g\in G$
be an element of maximal order. (In choosing g, we have already used the finiteness of G.) Observe that cliques in
$\Gamma _{\mathrm {EP}}(G)$
are in one-to-one correspondence with maximal cyclic subgroups of G (see [Reference Cameron2, Proposition 2.4]). It follows that independent sets in
$\mathcal {C}\Gamma _{\mathrm {EP}}(G)$
are in one-to-one correspondence with maximal cyclic subgroups of G. In particular,
$\langle g \rangle $
is an independent set of maximal size in
$\mathcal {C}\Gamma _{\mathrm {EP}}(G)$
, and hence, all the elements
$h\in G$
which are not powers of G are adjacent to g.
Let
$g^\ell $
be a power of g and suppose that
$g^\ell $
is not an isolated vertex (that is, it is not contained in every maximal cyclic subgroup of G). Then, there exists a maximal cyclic subgroup
$C\le G$
such that
$g^\ell \notin C$
. If we denote by
$h_\ell $
a generator of C,
$g^\ell $
and
$h_\ell $
are adjacent. Therefore,
$g^\ell $
is part of the connected component that contains all the
$h\in G$
which are not powers of g. This proves that
$\mathcal {C}\Gamma _{\mathrm {EP}}(G)$
, outside of its isolated vertices, has exactly one connected component.
We now focus on the upper bound on the diameter of the connected component of
$\mathcal {C}\Gamma _{\mathrm {EP}}(G)$
. Observe that our proof already shows that the diameter is bounded from above by
$4$
, and the vertices that might meet the maximal distance are distinct powers of g. Aiming for a contradiction, we suppose that, for some integers a and b,
$g^a$
and
$g^b$
are at distance
$4$
. We suppose that
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20250210114958251-0855:S0004972724001382:S0004972724001382_eqnu1.png?pub-status=live)
is a path of minimal distance connecting
$g^a$
and
$g^b$
. We point out that for
$i\in \{a,b\}$
,
$h_i$
is the vertex adjacent to
$g^i$
built in the previous paragraph. By minimality of this path,
$h_a$
and
$h_b$
are not adjacent. By the definition of
$\mathcal {C}\Gamma _{\mathrm {EP}}(G)$
,
$\langle h_a, h_b \rangle $
is a cyclic subgroup of G and, for
$i\in \{a,b\}$
, by construction,
$h_i$
generates a maximal cyclic subgroup of G. Therefore,
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20250210114958251-0855:S0004972724001382:S0004972724001382_eqnu2.png?pub-status=live)
It follows that
$g^a$
is adjacent to
$h_b$
and
$g^b$
is also adjacent to
$h_a$
. Hence,
$g^a$
and
$g^b$
are at distance
$2$
, which is a contradiction. This proves that the diameter of
$\mathcal {C}\Gamma _{\mathrm {EP}}(G)$
is at most
$3$
.
3 Proof of Corollary 1.2
The adjective enhanced suggest that there exists an original version of the power graph. Indeed, for a finite group G, the power graph
$\Gamma _{\mathrm {P}}(G)$
is defined as the graph on the vertex set G such that two elements are declared adjacent if one is a power of the other. We denote its complement by
$\mathcal {C}\Gamma _{\mathrm {P}}(G)$
.
Remark 3.1 [Reference Chakrabarty, Ghosh and Sen3, Theorem 2.12].
For any prime p and for any positive integer m,
$\Gamma _{\mathrm {P}}(C_{p^m})$
is complete. The converse is also true: if
$\Gamma _{\mathrm {P}}(G)$
is complete, then G is a cyclic p-group. Indeed, this follows from the fact that the series
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20250210114958251-0855:S0004972724001382:S0004972724001382_eqnu3.png?pub-status=live)
contains all the subgroups of
$C_{p^m}$
. Hence,
$\mathcal {C}\Gamma _{\mathrm {P}}(C_{p^m})$
contains no edges.
Proof of Corollary 1.2.
By construction,
$\Gamma _{\mathrm {P}}(G)$
is a subgraph of
$\Gamma _{\mathrm {EP}}(G)$
, and hence the complement of the power graph
$\mathcal {C}\Gamma _{\mathrm {P}}(G)$
contains
$\mathcal {C}\Gamma _{\mathrm {EP}}(G)$
as a subgraph. We need to focus our attention on those isolated vertices of
$\mathcal {C}\Gamma _{\mathrm {EP}}(G)$
that have neighbours in
$\mathcal {C}\Gamma _{\mathrm {P}}(G)$
. In the proof of Theorem 1.1, we have shown that all isolated vertices in
$\mathcal {C}\Gamma _{\mathrm {EP}}(G)$
are powers of an element of maximal order g.
Aiming for a contradiction, we suppose that there is an integer
$\ell $
such that
$g^\ell $
is contained in a connected component of
$\mathcal {C}\Gamma _{\mathrm {P}}(G)$
that does not contain g. It follows that
$g^\ell $
is not adjacent to any
$h \in G - \langle g \rangle $
. In particular,
$g^\ell $
is a power of every element of the group that is not a power of g. If two distinct primes divide the order of G, say p and r, then
$g^\ell $
lies both in a Sylow p-subgroup and a Sylow r-subgroup, because
$g^\ell $
is a power of both a p-element and an r-element. The only element with this property is the identity, and thus
$g^\ell =1$
is an isolated vertex. Therefore, G is a p-group, for a suitable prime p, and the neighbourhood of
$g^\ell $
is contained in
$\langle g \rangle $
. Since g is a p-element, by Remark 3.1,
$\mathcal {C}\Gamma _{\mathrm {P}}(\langle g \rangle )$
has no edges. Hence,
$g^\ell $
is an isolated vertex, which is a contradiction. This shows that
$\mathcal {C}\Gamma _{\mathrm {P}}(G)$
, outside of its isolated vertices, has a single connected component.
We now focus on the diameter of
$\mathcal {C}\Gamma _{\mathrm {P}}(G)$
. Since
$\mathcal {C}\Gamma _{\mathrm {P}}(G)$
is a subgraph of
$\mathcal {C}\Gamma _{\mathrm {EP}}(G)$
, Theorem 1.1 implies that if two vertices are at distance
$4$
in
$\mathcal {C}\Gamma _{\mathrm {P}}(G)$
, then they are distinct powers of g. Aiming for a contradiction, suppose that, for two suitable integers a and b,
$g^a$
and
$g^b$
are at distance
$4$
in
$\mathcal {C}\Gamma _{\mathrm {P}}(G)$
. Let
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20250210114958251-0855:S0004972724001382:S0004972724001382_eqnu4.png?pub-status=live)
be a path of minimal distance connecting
$g^a$
and
$g^b$
. By definition of
$\mathcal {C}\Gamma _{\mathrm {P}}(G)$
and by the minimality of the path, we have
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20250210114958251-0855:S0004972724001382:S0004972724001382_eqnu5.png?pub-status=live)
It might happen that
$z \in \langle x \rangle $
is true instead: in this case, we just swap a with b and x with z. Therefore,
$\langle x \rangle $
is a subgroup of
$\langle z \rangle $
. Hence,
$g^b$
cannot be an element of
$\langle x \rangle $
, and
$g^b$
and x are adjacent, against the minimality of the chosen path. This final contradiction completes our proof and our note.
We observe that the above proof actually never uses the assumption that
$g^a$
and
$g^b$
are powers of g.
Acknowledgement
The first author is a member of the GNSAGA INdAM research group and kindly acknowledges its support.