Further study on forbidden subgraphs of power graph

Let $G$ be a finite nilpotent group such that $P(G)$ is $P_{5}$-free.
Claim 1. $o(G)$ has at most two distinct prime divisors.
Proof of Claim 1: Suppose, $p,q,r$ are $3$ distinct primes divide $o(G)$. Let $a,b,c$ be the elements of order $p,q,r$ respectively. Then $P(G)$ contains a path $a\sim ab\sim b\sim bc\sim c$.
According to the above claim we have either $G$ is a $p$-group or $o(G)=p^{a}q^{b}$, where $p,q$ are distinct primes and $a,b\geq 1$. We now consider the following cases.
Case 1. Let $G$ be a $p$-group.
By Theorem 2.1, $P(G)$ is a cograph implies $P(G)$ is $P_{5}$-free.
Case 2. Let $o(G)$ has two distinct prime divisors say $p,q$.
Suppose, $o(G)=p^{a}q^{b}$ with $a,b\geq 1$. Let $P,Q$ be the Sylow $p$- and Sylow $q$-subgroups of $G$.
Claim 2. We claim that both Sylow subgroups must be cyclic.
For the sake of contradiction, let the Sylow $p$-subgroup $P$ be non-cyclic. Then there exist elements say $a,b$ of order $p$ such that $a\nsim b$ in $P(G)$. In that case, $P(G)$ contains a path $a\sim ac\sim c\sim bc\sim b$, where $c\in Q$. Therefore, $G\cong C_{p^{a}q^{b}}$ with $a,b\geq 1$.
If both $a,b>1$ then again the path $a\sim b\sim c\sim d\sim e$ is contained in $P(G)$, where $o(a)=p^{a},o(b)=p,o(c)=pq,o(d)=q,o(e)=q^{b}$. Thus one of $a$ or $b$ must be $1$ and hence $G$ is the cyclic group $C_{p^{a}q}$ with $a\geq 1$.
Converse part is obvious. ∎

Let $G$ be a finite nilpotent group such that $P(G)$ is $\{P_{5},\overline{P_{5}}\}$-free. Since, $P(G)$ is $P_{5}$-free so by Theorem 4.1 $G$ must be the groups i) or ii).
Converse Part:
i) Let $G$ be a $p$-group. If $P(G)$ contains $\overline{P_{5}}$ then $P(G)$ must comprises a $4$-vertex induced path. This contradicts the Theorem 2.1, that is $P(G)$ is a cograph.
ii) On the other hand, let $G$ be the cyclic group $C_{p^{a}q}$ ($a,b\geq 1$). Suppose $P(G)$ has induced subgraph $\overline{P_{5}}$. We choose $4$ consecutive vertices of orders $p^{a},p,pq,q$. Then the fifth vertex must be of order $p^{i}q$, which is adjacent to both the second and third vertices. This leads to a contradiction that $\overline{P_{5}}$ is an induced subgraph.
Thus, in any cases $P(G)$ is $\{P_{5},\overline{P_{5}}\}$-free. ∎

Suppose $P(G\times H)$ is $\{P_{5},\overline{P_{5}}\}$-free.
First observe that $o(G\times H)$ has at most two distinct prime divisors. Elsewhere there exists $3$ primes say $p,q,r$ such that $p,q|o(G)$ and $r|o(H)$. Then $P(G\times H)$ contains a path $a\sim ac\sim c\sim bc\sim b$, where $o(a)=p,o(b)=q$ and $o(c)=r$.
If $G$ and $H$ are both power of same prime then $G\times H$ is a $p$-group. So, there is nothing to prove (see Theorem 4.2).
Let $o(G\times H)$ has precisely two distinct prime divisors say $p,q$.
If both $G,H$ are abelian then $G,H$ have the structure in b).
Let us assume that both $G,H$ can not be abelian.
Claim: one of $o(G),o(H)$ must be of prime power order.
Proof of the claim Let $p,q$ be two primes such that $pq|o(G),o(H)$ both. Consider the Sylow $p$- and Sylow $q$-subgroups of both $G,H$ are $P_{G},Q_{G}$ and $P_{H},Q_{H}$ respectively. Now $P_{G}\times Q_{H}$ is nilpotent and $\{P_{5},\overline{P_{5}}\}$-free. Without loss of generality let $P_{G}\cong C_{p^{a}}$ and $Q_{H}\cong C_{q}$ ( by Theorem 4.2). Similarly, $Q_{G}\times P_{H}$ is nilpotent and $\{P_{5},\overline{P_{5}}\}$-free implies $Q_{G}\cong C_{q^{b}}$ ($b\geq 1$) and $P_{H}\cong C_{p}$ or $Q_{G}\cong C_{q}$ and $P_{H}\cong C_{p^{k}}$ ($k\geq 1$). Also, none of $G,H$ contains any abelian subgroup say $M$ of order $pq$; otherwise $G\times H$ contains one of the four nilpotent subgroups $P_{G}\times M$, $Q_{G}\times M$, $M\times P_{H}$, $M\times Q_{H}$, whose power graphs comprise a $P_{5}$ [see Theorem 4.2].
Since, $G,~H$ can not have abelian subgroup of order $pq$ so they are not nilpotent. So, one of the two Sylow subgroups of them must be not normal. Without loss of generality, let the Sylow $q$-subgroups of $G$ be not normal. Then the Sylow $p$-subgroup $P_{H}$ must be normal as otherwise $P(G\times H)$ has $P_{5}$. Using the similar argument we can say that the Sylow $q$-subgroup of $H$ ($Q_{H}$) is not normal and the Sylow $p$-subgroup of $G$ ($P_{G}$) is normal (since both $G,H$ are not nilpotent). Hence $G\cong C_{p^{a}}\rtimes C_{q}$ and $H\cong C_{p^{k}}\rtimes C_{q}$ or $G\cong C_{p^{a}}\rtimes C_{q^{b}}$ and $H\cong C_{p}\rtimes C_{q}$. But in any of the cases we obtain that $G$ contains two elements of order $q$ (say, $a,b$) such that $a\nsim b$ in $P(G\times H)$ and $H$ contains an element of order $p$, say $c$, for which $P(G\times H)$ has a path $a\sim ac\sim c\sim bc\sim b$.
Therefore, one of $G,H$ must be the group of prime power order. Let $G$ be the group with $o(G)=q^{m}$ and $pq|o(H)$.
Clearly, $G$ must be cyclic elsewhere $G$ contains two elements of order $q$ that are non-adjacent in $P(G\times H)$ and $H$ has an element of order $p$ such that $P(G\times H)$ comprises a $5$-vertex induced path.
Let $P_{H},Q$ be Sylow $p$- and Sylow $q$-subgroups of $H$. Now, $G\times P_{H}$ is nilpotent gives either $G\cong C_{q}$ and $P_{H}\cong C_{p^{r}}$ or $G\cong C_{q^{m}}$ and $P_{H}\cong C_{p}$. But clearly $H$ does not consist any abelian subgroup $K$ of order $pq$ as otherwise $G\times H$ contains a nilpotent subgroup $G\times K$ whose power graph has $P_{5}$ (by Theorem 4.2). This implies that either $G\cong C_{q}$ and $H\cong C_{p^{r}}\rtimes Q$ or $G\cong C_{q^{m}}$ and $H\cong C_{p}\rtimes Q$, where $Q$ is the Sylow $q$-subgroup of $H$. Thus we get the structures of $G,H$ which take the form as given in c).
Converse Part:
If $G$ and $H$ are either a) or b) then $P(G\times H)$ is $\{P_{5},\overline{P_{5}}\}$-free by Theorem 4.2.
Let $G$ and $H$ have the form as in c). Firstly, we prove that $P(G)$ is $P_{5}$-free. If possible let $P(G)$ contain a $5$-vertex induced path $P_{5}$. Let $x_{1}\rightarrow x_{2}\leftarrow x_{3}\rightarrow x_{4}\leftarrow x_{5}\rightarrow\cdots$ be such a path, where $x_{i}=(g_{i},h_{i})$.
If $o(h_{3})=1$, then $x_{2}\sim x_{4}$ as $G$ is cyclic.
If $o(g_{3})=1$, then again $x_{2}\sim x_{4}$ since $H$ contains only elements of order power of either $p$ or $q$.
Let $o(g_{3})$ and $o(h_{3})$ both be a power of $q$. Then we obtain that $x_{2}\sim x_{4}$.
Suppose $o(g_{3})$ is a power of $q$ and $o(h_{3})$ is a power of $p$. Let $g_{3}=a,h_{3}=b$. Without loss of generality, we assume that $g_{2}=1,h_{2}=b^{q^{k}}$ and $g_{4}=a^{p^{l}},h_{4}=1$. Then as $x_{1}\sim x_{2}$ and $G$ is cyclic so $x_{1}\sim x_{3}$. Similarly, if we reverse $(g_{2},h_{2})$ and $(g_{4},h_{4})$ then we have either $x_{1}\sim x_{3}$ or $x_{3}\sim x_{5}$. Thus, $P(G)$ does not contain $P_{5}$.
Next we prove that $P(G)$ is $\overline{P_{5}}$-free. Clearly, if a graph contains $\overline{P_{5}}$ then the graph must have an induced $4$-vertex cycle. Suppose, $P(G)$ carries $\overline{P_{5}}$. Then $P(G)$ has a $4$-vertex induced cycle say $A\sim B\sim C\sim D\sim A$ with $C=(a,b)$. If one of $a$ or $b$ is $1$ then $B\sim D$. On the other hand, if $a,b$ are both power of $q$ then also $B\sim D$; whereas for the case $o(a)=q^{k},o(b)=p^{l}$ we obtain $A\sim C$. Thus, $P(G)$ does not contain any $4$-vertex induced cycle and hence $P(G)$ is $\overline{P_{5}}$-free. ∎

If $n\geq 6$ then $P(S_{n})$ contains a path $(5~6)\sim(1~2~3)(5~6)\sim(1~2~3)\sim(1~2~3)(4~5)\sim(4~5)\sim(4~5)(1~2~6)$. So, $n\leq 5$.
If $n\leq 5$ then the maximal cyclic subgroups of $P(S_{n})$ intersect in the identity, and their orders are in the set $\{2\}$ (for $n=2$), $\{2,3\}$ (for $n=3$), $\{2,3,4\}$ (for $n=4$), or $\{4,5,6\}$ (for $n=5$). As for each $n\leq 5$, the power graph of the maximal cyclic subgroups of $S_{n}$ does not contain any path of length $3$ and above so $P(S_{n})$ is $P_{5}$-free and $\overline{P_{5}}$-free (since, $\overline{P_{5}}$ contains induced $P_{4}$). ∎

Firstly consider the Mathieu group $M_{11}$. It contains $165$ elements of order $2$, $440$ elements of order $3$, $1320$ elements of order $6$ and $990$ elements of order $4$. So, there exist elements $a,b,c,d,e$ of orders $4,2,6,3,6$ resp. such that $a^{2}=b=c^{3},c^{2}=d=e^{2}$ with $c^{3}\neq e^{3}$. Thus, $P(M_{11})$ contains $P_{5}$.
Since, $M_{11}$ is contained as a subgroup in every sporadic group except $J_{1},J_{2},J_{3},M_{22},He,Ru$ and $Th$ so their power graph contains $P_{5}$.
But $M_{22}$ contains $A_{7}$, $J_{1},J_{2},J_{3}$ contain $D_{3}\times D_{5},A_{4}\times A_{5},C_{3}\times A_{6}$ respectively, $He,Ru$ contain $S_{7},A_{8}$ respectively. By Theorems 4.4, 4.7, 4.3 the power graphs of these subgroups contains either $P_{5}$ or its complement. Hence, the power graphs of these groups are not $\{P_{5},\overline{P_{5}}\}$-free. From the information in $\mathbb{ATLAS}$ [7], we observe that $Th$ contains elements $x,y,z,w,u$ whose orders are $3,6,2,10,5$ respectively. Additionally these elements satify the conditions $y^{2}=x,y^{3}=z=w^{5}$ and $u=w^{2}$. Thus, $P(Th)$ contains a $5$ vertex induced path $x\sim y\sim z\sim w\sim u$.
This completes the proof of the theorem. ∎

For $n\geq 7$, $P(A_{n})$ contains a path $(1~2~3~4)(5~6)\sim(1~3)(2~4)\sim(1~3)(2~4)(5~6~7)\sim(5~6~7)\sim(1~2)(3~4)(5~6~7)$. Thus, $n\leq 6$.
For $n=4,5,6$ then prime graph of $A_{n}$ is a null graph; so $P(A_{n})$ must be $P_{5}$-free. Otherwise, $P(A_{n})$ contains a $4$-vertex induced path which contradicts that $P(A_{n})$ is a cograph (see Theorem 2.2). Since, $\overline{P_{5}}$ contains induced path $P_{4}$ and $P(A_{n})$ (where $n=4,5,6$) is $P_{4}$-free so $P(A_{n})$ is also $\overline{P_{5}}$-free.
If $n=3$ then $P(A_{3})$ is a complete graph (as $A_{3}$ is a cyclic group of order $3$). This implies that $P(A_{3})$ is $\{P_{5},\overline{P_{5}}\}$-free. ∎

Let $q$ be a power of some odd prime. Now, $P^{*}(G)$ is the disjoint union of $P^{*}(C_{(q\pm 1)/2})$ along with some isolated vertices. Thus if $P(G)$ is $\{P_{5},\overline{P_{5}}\}$-free then $P^{*}(C_{(q\pm 1)/2})$ is also $\{P_{5},\overline{P_{5}}\}$-free. This implies the condition in a) according to the Theorem 4.2.
Suppose $q$ is a power of $2$. Then $P^{*}(G)$ is the disjoint union of $P^{*}(C_{(q\pm 1)})$ along with some isolated vertices. Then by Theorem 4.2 the numbers $q\pm 1$ satisfy the conditions in b). ∎

Here $G$ has $4$ maximal cyclic subgroups of orders $4,q-1,q\pm\sqrt{2q}+1$. Since these $4$ numbers are pairwise coprime so any edge in $P(G)$ must lie in a maximal cyclic subgroup. Thus if $P(G)$ contains either $P_{5}$ and its complement then it must be contained in a maximal cyclic subgroup. Now the power graph of a cyclic group of order $4$ is a complete graph. Therefore, $P(G)$ is $\{P_{5},\overline{P_{5}}\}$-free if and only if the numbers $q-1,q\pm\sqrt{2q}+1$ satify the stated condition according to the Theorem 4.2. ∎

Let $\beta$ be a generator of the multiplicative group of $GF(q^{2})$. So $o(\beta)=q+1$. Let $p(>3)$ be a prime factor of $q+1$. Set $d=(q+1)/p$. Now $\alpha=\beta^{d(q-1)}$ has order $p$. Then $\overline{\alpha}=\beta^{d(q^{2}-1)}$ and $\alpha\overline{\alpha}=1$ in $GF(q^{2})$. Choose $3$ matrices $g,h,k$ as follows:

Then $o(g)=2,o(h)=p,o(k)=o(gh)=o(hk)=2p$ and $k^{2}=h$, $gh=hg$, $hk=kh$. So, $P(SU(3,q))$ contains the induced path $g\sim gh\sim h\sim k\sim k^{p}$. Since $g,h,k\in SU(3,q)\setminus Z$ so choosing $a=gZ,b=hZ,c=kZ$; then an induced path $a\sim ab\sim b\sim c\sim c^{p}$ is contained in $P(PSU(3,q))$.
But this argument is not valid when $q=8$. In that case, $P(PSU(3,8))$ contains a subgroup $C_{3}\times PSL(2,8)$ whose power graph is not $\{P_{5},\overline{P_{5}}\}$-free (by Theorem 4.3). ∎

If $q$ is odd, then $PSU(3,q)$ contains a cyclic subgroup of order $(q^{2}-1)/gcd(q+1,3)$. Since $q$ is odd, both the numbers $q-1$ and $(q+1)/gcd(q+1,3)$ are even. Thus $P(PSU(3,q))$ is $\{P_{5},\overline{P_{5}}\}$-free if $(q-1)(q+1)/gcd(q+1,3)$ is either a power of $2$ or of the form $2^{k}p^{\prime}$, where $p^{\prime}$ is an odd prime.
First suppose that both $(q+1)/gcd(q+1,3)$ and $q-1$ are powers of $2$. As only one of $q+1$ and $q-1$ is divisible by $4$, so the pair $(q-1,q+1)$ is either $(2,4)$ or $(4,6)$. Hence $q=3$ or $5$.
Next, suppose that $(q-1)(q+1)/gcd(q+1,3)=2^{k}p^{\prime}$. Then one of $q-1,(q+1)/gcd(q+1,3)$ is a power of $2$. Without loss of generality, we assume that $q-1$ is a power of $2$. Now if $q\neq 3,9$, then $q+1$ is either of the forms $2p^{\prime}$ or $6p^{\prime}$ for some odd prime $p^{\prime}$. If $q+1=6p^{\prime}$, then $PSU(3,q)$ contains a subgroup $C_{2^{k}}\times C_{2p}$ whose power graph is not $\{P_{5},\overline{P_{5}}\}$-free (by Theorem 4.2). Again, if $q+1=2p^{\prime}$, then $PSU(3,q)$ contains the subgroup $C_{q+1}\times C_{q+1/gcd(q+1,3)}$. By Theorem 4.2, $P(C_{2p^{\prime}}\times C_{2p^{\prime}})$ is not $\{P_{5},\overline{P_{5}}\}$-free. So either $q=3$ or $9$ in this case.
But if $q=9$ then $C_{8}\times C_{10}$ is contained in $PSU(3,q)$. For $q=5$, $PSU(3,5)$ contains $A_{7}$. The power graph of none of these subgroups are $\{P_{5},\overline{P_{5}}\}$-free [see Theorem 4.2]. On the other hand, if $q=3$ then $PSU(3,3)$ contains $4\cdot S_{4}$. Now, $4\cdot S_{4}$ is given by $\textlangle a,b,c,d,e|a^{4}=d^{3}=1,b^{2}=c^{2}=e^{2}=a^{2},ab=ba,ac=ca,ad=da,eae^{-1}=a^{-1},cbc^{-1}=a^{2}b,dbd^{-1}=a^{2}bc,ebe^{-1}=bc,dcd^{-1}=b,ece^{-1}=a^{2}c,ede^{-1}=d^{-1}\textrangle$. Then $P(PSU(3,5))$ contains the induced path $a\sim ad\sim d\sim ed\sim e$. Thus, in any of the case $q=3,5,9$, the power graph of $PSU(3,q)$ is not $\{P_{5},\overline{P_{5}}\}$-free. ∎