Proof. It is clear that if an edge is present in the commuting graph, then it is present in the symmetric normaliser graph. Let x and y be adjacent in SNorm(G), and let H denote the subgroup of G generated by x and y. Since the commutator $c = [x,y] = xyx^{-1}y^{-1}$ commutes with both x and y, the subgroup generated by c is in the centre of H. Clearly, the quotient group ${H}/{\langle c \rangle }$ is abelian and, therefore, nilpotent. Consequently, H is nilpotent.

Now, let G be a nonabelian simple group and assume that $x,y$ are nonidentity elements. Then, $H \leq N(\langle y \rangle ) \lneq G$ , which implies that $x,y$ are adjacent in NGen(G). Lastly, since the commutator $[x, y]$ commutes with both x and y, we have $[x,_2 y ] = 1$ and $[y ,_2 x] =1$ . Therefore, elements $x,y$ are adjacent in the Engel graph of G.

Proof. Let G be a finite group such that the edge set of its commuting graph is a proper subset of the edge set of its symmetric normaliser graph. This gives elements x and y, which satisfy the relations

(2.2) $$ \begin{align} ab \neq ba, \quad a \in N(\langle b \rangle) \quad\mbox{and}\quad b \in N(\langle a \rangle). \end{align} $$

Note that if $x, y$ satisfy (2.2), then $[x, y]$ commutes with both x and y. So, using (2.1a) and (2.1b), we have $ [x^m, y]= {[x, y]}^m = [x, y^m]$ for any integer m.

Let $\mathcal {L}$ denote the collection of all elements $x'$ such that there exists $y' \in G$ with $x', y'$ satisfying (2.2). Choose h from $\mathcal {L}$ such that $|h|$ is the least. Define a set $\mathcal {L}_h$ as the collection of all elements $y"$ such that h and $y"$ satisfy (2.2). From this set, we choose k such that $|k|$ is the least. Let p be a prime that divides the order of h. Now, $h^p$ lies in the normaliser of $\langle k \rangle $ in G and k lies in the normaliser of $\langle h^p \rangle $ in G. This implies that $h^p$ commutes with k and so $[h,k]^p=e$ . Similarly, if q is a prime that divides the order of k, then $[h,k]^q=e$ , which means that p and q are equal. Thus, h and k have prime-power order for the same prime, say p. Let $H=\langle h,k \rangle $ . Then, the subgroup H has order $p^n$ for some natural number n, and its derived subgroup $H' = \langle [h,k] \rangle $ has order p and $H' \leq Z(H)$ , which makes H a $2$ -generated nilpotent group of class 2.

Proof. Consider the parameters in cases A1(b) and B1(b). Then, $\sigma = 1$ and $\rho = 1$ . Let $x=a^ib^j[a,b]^k$ and $y=a^rb^s[a,b]^t$ be two distinct elements of P such that they are adjacent in SNorm(P). We shall prove that they commute with each other, so they cannot generate the nonabelian group P. It follows by induction on m that

(2.4) $$ \begin{align} y^m = a^{rm}b^{sm}[a,b]^{tm-rsm(m-1)/2}. \end{align} $$

The elements x and y are adjacent in SNorm(P), so there must exist integers d and e such that

$$ \begin{align*}yxy^{-1}=x^d \quad\mbox{and}\quad xyx^{-1} =y^e.\end{align*} $$

Substituting values of $x,y$ and using $b^lab^{-l} = a[b,a]^l$ gives

(2.5) $$ \begin{align} xyx^{-1} = a^rb^s[b,a]^{rj-si-t}. \end{align} $$

Equating (2.5) and (2.4) for $m=e$ , we get $re \equiv r \mbox { mod } p^{\alpha }$ . If p does not divide r, we get $y^e=y$ and so $xyx^{-1}=y$ . Similarly, if p does not divide i, we have $x^d=x$ and so ${yxy^{-1}~=x}$ . If p divides both r and i, then p divides $rj-si$ and $[x,y] = [b,a]^{rj-si} =~1$ . Hence, x and y generate an abelian subgroup of P whenever they are adjacent in SNorm(P).

For cases A1(c) and B1(c), we have $\rho = 1$ and $\sigma = 0$ . Let $x,y \in P$ . Then, there exist unique $i,j,r,s$ satisfying $1 \leq i,r \leq p^{\alpha }$ and $1 \leq j,s \leq p^{\beta + 1}$ such that $x=a^ib^j$ and $y=a^rb^s$ . Let $x,y$ be adjacent in SNorm(P). We shall prove that they commute with each other. Since x and y are adjacent in SNorm(P), there must exist integers d and e such that

$$ \begin{align*}yxy^{-1}=x^d \quad\mbox{and}\quad xyx^{-1} =y^e.\end{align*} $$

Proceeding as in the previous case, we get $xy=yx$ .

Lastly, we consider cases A2(c), B2(c) and B3(b). In these cases, we have $(\alpha , \beta , \rho , \sigma ) = (\alpha , \alpha , 1, 1)$ with $\alpha \geq 1$ . If x and y are adjacent in SNorm(P), then $|\langle x,y \rangle | =~{|x||y|}/{|\langle x \rangle \cap \langle y \rangle |} \leq p^{2\alpha } = p^{n-1} < |P|$ .

Proof. We will prove this result by showing the existence of x and y, which satisfy the conditions in each of the SNNC-groups.

In case B3(a), when $(\alpha , \beta , \rho , \sigma ) = (1, 1, 0, 0)$ , the group P is isomorphic to the quaternion group $Q_8$ . The symmetric normaliser graph of $Q_8$ is a complete graph. So, $Q_8$ has 2 elements that generate it and are also adjacent in SNorm( $Q_8$ ).

For the parameters that do not give $Q_8$ , the proof is as follows. We have $ [a,b] = a^{p^\alpha }$ , which gives us $ba^{-1}b^{-1}=a^{p^{\alpha }-1}$ . Thus,

(2.6) $$ \begin{align} a(ab)a^{-1} = a^2(ba^{-1}b^{-1})b = a^2(a^{-1+p^{\alpha}})b=a^{1+p^{\alpha}}b. \end{align} $$

Since $\langle a \rangle $ is normal in P, we have $ab \in N(\langle a \rangle )$ . We claim that $a \in N(\langle ab \rangle )$ . Substituting $i=1+p^{\alpha }$ in (2.3),

(2.7) $$ \begin{align} \begin{aligned} (ab)^i &= [b,a]^{(1+p^{\alpha})(p^{\alpha}/2)}a^{1+p^{\alpha}}b^{1+p^{\alpha}}\\ &= a^{1+p^{\alpha}}b^{1+p^{\alpha}} \quad\text{(since } (1+p^{\alpha})p^{\alpha}/2 \equiv 0 \text{ mod } p \text{)}\\ &= a^{1+p^{\alpha}}b. \end{aligned} \end{align} $$

Equating (2.6) and (2.7),

$$ \begin{align*}a(ab)a^{-1} = a^{1+p^{\alpha}}b = (ab)^{1+p^{\alpha}} \in \langle ab \rangle.\end{align*} $$

This gives x and y that generate P and are adjacent in SNorm(P).

Proof. Let G be a finite group such that the two graphs of G are unequal. By Lemma 2.2, the group G has a subgroup $H = \langle h, k \rangle $ , which is a 2-generated p-group such that $|H'| =p$ and $H' = Z(H)$ . Further, $h, k$ are adjacent in SNorm(H). By [Reference Blackburn, Cocke, Misseldine and Venkataraman2, Theorem D], the subgroup H is an exponent-critical group of type $\mathcal {B}$ . By Lemma 2.3, the subgroup H is isomorphic to one of the groups whose parameters are mentioned in the list. Since the subgroup H has elements h and k that generate it and are adjacent in its symmetric normaliser graph, by Lemma 2.4 and Lemma 2.5, the subgroup H must be an SNNC-group.

Conversely, if a finite group G has a subgroup isomorphic to an SNNC-group, then G will have elements that satisfy (2.2), which means that the commuting graph of G is not equal to the symmetric graph of G.

Proof. Let G be a finite group such that the symmetric normaliser graph and the nilpotent graph of G are equal. Let P denote a Sylow p-subgroup of G. Since the nilpotent graph of P is complete, the symmetric normaliser graph of P is also complete, which makes the subgroup P a Dedekind group. By [Reference Robinson8, Theorem 5.3.7], a group is Dedekind if and only if it is either abelian or the direct product of a quaternion group of order 8, an elementary abelian 2-group and an abelian group with all its elements of odd order. Thus, for odd primes p, the Sylow p-subgroups are abelian, and the Sylow $2$ -subgroups of G are either abelian or a direct product of $Q_8$ and an elementary abelian $2$ -group.

Conversely, for the finite group G and for an odd prime p, let the Sylow p-subgroups be abelian and for $p=2$ , let the Sylow $2$ -subgroups either be abelian or a direct product of $Q_8$ and an elementary abelian $2$ -group. Let x and y be adjacent in Nilp(G). Since $H=\langle x,y \rangle $ is a subgroup of G, for an odd prime p, the Sylow p-subgroups of H are abelian and for $p=2$ , the Sylow $2$ -subgroups of H are either abelian or a direct product of $Q_8$ and an elementary abelian $2$ -group. Since H is nilpotent, it is a direct product of its Sylow subgroups. By [Reference Robinson8, Theorem 5.3.7], the subgroup H is a Dedekind group and its symmetric normaliser graph is complete. Thus, x and y are adjacent in SNorm(G).

Proof. Let G be a finite group such that its enhanced power graph and symmetric normaliser graphs are the same. This implies that EPow(G) = Com(G) = SNorm(G). By [Reference Cameron3, Proposition 3.2], Sylow p-subgroups of G are cyclic or generalised quaternion groups. Since the quaternion group is an SNNC group, by Theorem 2.6, the Sylow p-subgroups of G must be cyclic. Conversely, if the Sylow p-subgroups of G are cyclic, then G does not have a subgroup that is an SNNC group. Thus, Com(G) = SNorm(G) and, by [Reference Cameron3, Proposition 3.2], EPow(G) = Com(G).

Question 3.3. What are the finite groups G of order $p^{\alpha }q^{\beta }$ for distinct primes $p,q$ that are Frobenius or $2$ -Frobenius and have cyclic Sylow subgroups? If such a G is Frobenius, it will be a semidirect product of a Frobenius kernel K isomorphic to ${\mathbb {Z}_{p^{\alpha }}}$ and a Frobenius complement H isomorphic to ${\mathbb {Z}_{q^{\beta }}}$ .

Question 3.4. Let G be a nonabelian simple group. When does SNorm(G) = NGen(G)? Or is it always the case that SNorm(G) $\subsetneq $ NGen(G)? From [Reference Cameron3, Proposition 3.1(c)], for any group H, we have NGen(H) = Com(H) if and only if H is abelian and not $2$ -generated or a minimal nonabelian group. Since a minimal nonabelian group is always solvable, we see that for a finite nonabelian simple group G, we always have E(Com(G)) $\subsetneq \, E$ (NGen(G)). Also, any SNNC-group P for an odd prime p is always contained in an $A_n$ for all sufficiently large n. This gives E(Com(G)) $\subsetneq \, E$ (SNorm(G)) when $G = A_n$ for all n suitably large.

Question 3.5. What are the necessary and sufficient conditions for SNormG to be equal to Engel(G)? If G is a Dedekind group, then SNorm(G) = Nilp(G) = Engel(G), as all are complete graphs. However, we know that if the Sylow subgroups of G are Dedekind, then SNorm(G) = Nilp(G).

Question 3.6. Let G be a non-Dedekind finite group whose Sylow subgroups are Dedekind. Is E(Nilp(G)) $\subsetneq \, E$ (Engel(G))? Note that for $G = S_3$ , we have E(Pow(G)) = E(EPow(G)) = E(Com(G)) = E(SNorm(G)) = E(Nilp(G)) $\subsetneq \, E$ (Engel(G)).