Group-Annihilator Graphs Realised by Finite Abelian Groups and Its Properties

Abstract

Let G be a finite abelian group viewed a ${\mathbb {Z}}$-module and let ${\mathcal {G}} = (V, E)$ be a simple graph. In this paper, we consider a graph $\Gamma (G)$ called as a group-annihilator graph. The vertices of $\Gamma (G)$ are all elements of G and two distinct vertices x and y are adjacent in $\Gamma (G)$ if and only if $[x : G][y : G]G = \{0\}$, where $x, y\in G$ and $[x : G] = \{r\in {\mathbb {Z}} : rG \subseteq {\mathbb {Z}}x\}$ is an ideal of a ring ${\mathbb {Z}}$. We discuss in detail the graph structure realised by a group G. Moreover, we study the creation sequence, hyperenergeticity and hypoenergeticity of group-annihilator graphs. Finally, we conclude the paper with a discussion on Laplacian eigen values of the group-annihilator graph. We show that the Laplacian eigen values are representatives of orbits of the group action: $Aut(\Gamma (G)) \times G \rightarrow G$.

Appendix: Proof of Theorem 3

We furnish the proof of Theorem 3 in this section.

Proof

• Consider the case $a \in {\mathcal {O}}_{\alpha , p^{\alpha }}$, and $b\in {\mathcal {O}}_{\beta , p^{\beta }}$

  • For $c \in {\mathcal {O}}_{\gamma , p^{\gamma }},$ it is trivially true that $[(a,b,c):G]= p^{\gamma }{\mathbb {Z}}.$

  • Let $c \in {\mathcal {O}}_{\gamma , p^i}$, where $0\le i \le \beta -1.$ Clearly, $p^{\beta }{\mathbb {Z}} \subset [(a,b,c): G].$ Now by considering $y \in [(a,b,c):G]$, we see that $y (0 ,1, 0)= n (0,0,c)$ for some integer n. That completes the other part.

  • Let $c \in {\mathcal {O}}_{\gamma , p^i}$, where $\beta \le i \le \gamma -1.$ Then $c= p^i k'$, where $(k',p) = 1.$ Clearly, $p^{i}{\mathbb {Z}} \subset [(a,b,c): G].$ Now by considering $y \in [(a,b,c):G]$, we again see that $y (0 ,0, 1)= n (0,0,c)$ for some integer n. So, we are done.

• Consider the case $a \in {\mathcal {O}}_{\alpha , p^{\alpha }}$, so $a=0$ and $b\in {\mathcal {O}}_{\beta , p^{j}}$, where $0 \le j \le \beta -1.$ So, $b= p^jb'$ for some $b'$, where $(b',p)=1.$

  • For $c \in {\mathcal {O}}_{\gamma , p^{\gamma }},$ it is trivially true that $[(a,b,c):G]= p^{\gamma }{\mathbb {Z}}.$

  • Let $c \in {\mathcal {O}}_{\gamma , p^i}$, where $0\le i \le j$. So, $c= p^ic'$ for some $c'$, where $(c',p)=1.$ Then for any $y = p^{\beta }k \in p^{\beta }{\mathbb {Z}},$ $y (u,v,w) = (0, 0, p^\beta ) kw = wkc'^{-1}p^{\beta -i} (a,b,c)$ as $j-i \ge 0.$ So $p^{\beta }{\mathbb {Z}} \subset [(a,b,c): G].$ Now by considering $y \in [(a,b,c):G]$, we have $y (0 ,1, 0)= n (0,0,c)$ for some integer n,  so we are done, since $j-i \ge 0.$

  • Let $c \in {\mathcal {O}}_{\gamma , p^i}$, where $j+1\le i \le \gamma - \beta +j.$ Then $c= p^i c'$ for some $c'$, where $(c',p) = 1.$ For any $y = p^{i+\beta -j}k' \in p^{\beta }{\mathbb {Z}},$ $y (u,v,w) = (0, 0, p^{\beta +i-j}) k'w = k'wc'^{-1}p^{\beta -j} (a,b,c)$ as $\beta - j \ge 1.$ So, $p^{\beta +i -j}{\mathbb {Z}} \subset [(a,b,c): G].$ Now by considering $y \in [(a,b,c):G]$, we have $y (0 ,0, 1)= n (0,b,c)$ for some integer n,  this implies $p^{\beta -j}\mid n$ and $y = p^{\beta +i-j}c' (\bmod p^{\gamma } )$, hence the result follows.

  • Let $c \in {\mathcal {O}}_{\gamma , p^i}$, where $\gamma - \beta +j+1\le i \le \gamma - 1.$ Then $c= p^i c'$ for some $c'$, where $(c',p) = 1.$ Clearly $p^{\gamma }{\mathbb {Z}} \subset [(a,b,c): G].$ Now by considering $y \in [(a,b,c):G]$, we have $y (0 ,0, 1)= n (0,b,c)$ for some integer n. This implies $p^{\beta -j}\mid n$ and since $i+\beta - j \ge \gamma +1$, so $y = p^{\beta +i-j}c' (\bmod p^{\gamma } )$.

• Consider the case $a \in {\mathcal {O}}_{\alpha , p^k}$. Assume that $a= p^ka'$ for some $a'$, where $(a',p)=1$ for $0\le k \le \alpha -1$ and $b\in {\mathcal {O}}_{\beta , p^{\beta }},$

  • For $c \in {\mathcal {O}}_{\gamma , p^{\gamma }},$ it is trivially true that $[(a,b,c):G]= p^{\gamma }{\mathbb {Z}}.$

  • Let $c \in {\mathcal {O}}_{\gamma , p^i}$, where $0\le i \le \beta - \alpha +k$. So $c= p^ic'$ for some $c'$, where $(c',p)=1.$ For any $y = p^{\beta }k' \in p^{\beta }{\mathbb {Z}},$

    $$\begin{aligned}&y (u,v,w) = (0, 0, p^\beta ) k'w \\&\quad = k'wc'^{-1}(p^{\beta -i+k}a', 0 , p^{\beta -i} p^i c') \in {\mathbb {Z}}(a,0,c), \end{aligned}$$

    since $\beta -i+k \ge \alpha .$ So, $p^{\beta }{\mathbb {Z}} \subset [(a,b,c): G].$ Now by considering $y \in [(a,b,c):G]$, we have $y (0 ,1, 0)= n (a,0,c)$ for some integer n, so we get $y\in p^{\beta }{\mathbb {Z}}$ and we are done.

  • Let $c \in {\mathcal {O}}_{\gamma , p^i}$, where $\beta - \alpha +k +1\le i \le \gamma - \alpha +k-1.$ Then $c= p^i c'$ for some $c'$, where $(c',p) = 1.$ For any $y = p^{i+\alpha -k}k' \in p^{i+\alpha -k}{\mathbb {Z}},$

    $$\begin{aligned} y (u,v,w)= & {} (0, 0, p^{\alpha +i-k}c') c'^{-1}k'w\\= & {} (p^{\alpha -k}p^ka',0,p^{\alpha -k} p^i c') c'^{-1}k'w \in {\mathbb {Z}}(a,b,c), \end{aligned}$$

    since $i+\alpha -k \ge \beta +1.$ So, $p^{\alpha +i -k}{\mathbb {Z}} \subset [(a,b,c): G].$ Now by considering $y \in [(a,b,c):G]$, we have $y (0 ,0, 1)= n (p^ka',0,p^ic')$ for some integer n. This implies $p^{\alpha -k}\mid n$ and $y = p^{\alpha - k +i}n' (\bmod p^{\gamma } )$ for some integer $n'$. Since $\alpha - k +i \le \gamma -1,$ so $y\in p^{\alpha - k +i} {\mathbb {Z}}.$

  • Let $c \in {\mathcal {O}}_{\gamma , p^i}$, where $\gamma - \alpha +k\le i \le \gamma - 1.$ Then $c= p^i c'$ for some $c'$, where $(c',p) = 1.$ Clearly $p^{\gamma }{\mathbb {Z}} \subset [(a,b,c): G].$

    Now by considering $y \in [(a,b,c):G]$, we have $y (0 ,0, 1)= n (a,0,c)$ for some integer n. This implies $p^{\alpha - k}\mid n$ and since $\alpha -k+i \ge \gamma$, so $y = p^inc' (\bmod p^{\gamma } )$.

• Consider the case $a \in {\mathcal {O}}_{\alpha , p^k}$. Assume that $a= p^ka'$ for some $a'$, where $(a',p)=1$ for $0\le k \le \alpha -1$ and $b\in {\mathcal {O}}_{\beta , p^{j}},$ so $b=p^jb'$ for some $b'$, where $(b',p)=1$ for $0 \le j \le \beta -1.$

  • For $c \in {\mathcal {O}}_{\gamma , p^{\gamma }},$ it is trivially true that $[(a,b,c):G]= p^{\gamma }{\mathbb {Z}}.$

  • Let $c \in {\mathcal {O}}_{\gamma , p^i}$, where $0\le i \le \gamma - \beta +j$. So $c= p^ic'$ for some $c'$, where $(c',p)=1.$ Note that $\beta -i \ge \alpha .$ Then for any $y = p^{\beta }k' \in p^{\beta }{\mathbb {Z}},$

    $$\begin{aligned} y (u,v,w)= & {} (0, 0, p^\beta ) k'w = k'c'^{-1}w (p^{\beta -i}p^k a',\\&p^{\beta -i}p^j b', p^{\beta -i} p^i c') \in {\mathbb {Z}}(a,b,c). \end{aligned}$$

    So $p^{\beta }{\mathbb {Z}} \subset [(a,b,c): G].$

    Now by considering $y \in [(a,b,c):G]$, we have $y (0 ,1, 0)= n (a,b,c)$ for some integer n. Then $p^{\gamma -i} \mid n$ and $y\equiv p^jc' (\bmod )$, so $y \in p^{\beta } {\mathbb {Z}}$, since $\gamma + (j-i) > \beta .$

  • Let $c \in {\mathcal {O}}_{\gamma , p^i}$, where $0\le i \le \gamma - \beta +j$. So, $c= p^ic'$ for some $c'$, where $(c',p)=1.$ Since $i> j$, so $\beta -j \ge \alpha .$ Then for any $y = p^{\beta +i -j}k' \in p^{\beta +i-j}{\mathbb {Z}},$

    $$\begin{aligned} y (u,v,w)= & {} (0, 0, p^{\beta +i-j}c') k' w= k' w(0, p^{\beta -j}p^jb', p^{\beta +i-j}c') \\= & {} k'w (p^{\beta -j}p^ka', p^{\beta -j}p^jb', p^{\beta +i-j}c') \in {\mathbb {Z}}(a,b,c) \end{aligned}$$

    as $i>j.$ So $p^{\beta }{\mathbb {Z}} \subset [(a,b,c): G].$

    Now by considering $y \in [(a,b,c):G]$, we have $y (0 ,0, 1)= n (a,b,c)$ for some integer n. Then $p^{\beta -j} \mid n$ and $y\equiv p^ic' (\bmod )$, so $y \in p^{\beta +i-j} {\mathbb {Z}}$, since $i< \gamma -\beta +j.$

  • Let $c \in {\mathcal {O}}_{\gamma , p^i}$, where $i \ge j.$ Then $c= p^i c'$ for some $c'$, where $(c',p) = 1.$ Also assume $i>\beta -\alpha >j.$ For any $y = p^{i+\beta -j}k' \in p^{i+\beta -j}{\mathbb {Z}},$

    $$\begin{aligned} y (u,v,w)= & {} (0,0,p^{\beta +i-j} c')c'^{-1}k'wa'\\= & {} (p^{\beta -j}p^ka',p^{\beta -j}p^jb',p^{\beta +i-j} c')k'wc'^{-1} \in {\mathbb {Z}}(a,b,c) \end{aligned}$$

    as $\beta -\alpha >j.$ So, $p^{\beta +i -j}{\mathbb {Z}} \subset [(a,b,c): G].$

    Now by considering $y \in [(a,b,c):G]$, we have $y (0 ,0, 1)= n (p^ka',p^jb',p^ic')$ for some integer n. This implies $p^{\beta -j}\mid n$ and $y = p^{i}nc' (\bmod p^{\gamma })$. Since $i < \gamma -\beta +j,$ so $y\in p^{\beta +i - j +} {\mathbb {Z}}.$.

  • Let $c \in {\mathcal {O}}_{\gamma , p^i}$, where $i \ge j.$ Then $c= p^i c'$ for some $c'$, where $(c',p) = 1.$ Also assume $\beta -\alpha < i$ , $i <\gamma -\alpha +k$ and $j> \beta -\alpha +k.$ For any $y = p^{i+\alpha -k}k' \in p^{i+\alpha -k}{\mathbb {Z}},$

    $$\begin{aligned} y (u,v,w)= & {} (0,0,p^{\alpha +i-k} c')k'wc'^{-1} \\= & {} (p^{\alpha -k}p^k a',p^{\alpha -k}p^jb',p^{\alpha +i-k} c')k'wc'^{-1} \in {\mathbb {Z}}(a,b,c) \end{aligned}$$

    as $\alpha -k > \beta -j.$ So, $p^{\alpha +i -k}{\mathbb {Z}} \subset [(a,b,c): G].$

    Now by considering $y \in [(a,b,c):G]$, we have $y (0 ,0, 1)= n (p^ka',p^jb',p^ic')$ for some integer n. This implies $p^{\alpha -k}\mid n$ and $y = p^{i}nc' (\bmod p^{\gamma })$. Since $i < \gamma - \alpha +k,$ so $y\in p^{\alpha +i - k} {\mathbb {Z}}.$

  • Let $c \in {\mathcal {O}}_{\gamma , p^i}$, where $i \ge j.$ Then $c= p^i c'$ for some $c'$, where $(c',p) = 1.$ Also assume $\beta -\alpha < i$ , $\gamma -\alpha +k \le i$ and $j> \beta -\alpha +k.$ From the previous argument $i \ge \gamma -\alpha +k$, therefore we have $p^{\gamma }{\mathbb {Z}}= [(a,b,c):G].$

  • Let $c \in {\mathcal {O}}_{\gamma , p^i}$, where $i \ge j.$ Then $c= p^i c'$ for some $c'$, where $(c',p) = 1.$ Also assume $\beta - \alpha < j \le \beta -\alpha +k.$ For any $y = p^{i+\beta -j}k' \in p^{i+\beta - j}{\mathbb {Z}},$

    $$\begin{aligned} y (u,v,w)= & {} (0,0,p^{\beta +i-j} c')k'wc'^{-1}\\= & {} (p^{\beta -j+k}a',p^{\beta -j}p^jb',p^{\beta +i-j} c')k'wc'^{-1} \in {\mathbb {Z}}(a,b,c) \end{aligned}$$

    as $\beta -\alpha < j \le \beta -\alpha +k.$ So, $p^{\beta +i -j}{\mathbb {Z}} \subset [(a,b,c): G].$

    Now by considering $y \in [(a,b,c):G]$, we have $y (0 ,0, 1)= n (p^ka',p^jb',p^ic')$ for some integer n. This implies $p^{\beta -j}\mid n$ and $y = p^{i}nc' (\bmod p^{\gamma })$. Since $i < \gamma - \beta +j,$ so $y\in p^{\beta +i - j} {\mathbb {Z}}.$

  • Let $c \in {\mathcal {O}}_{\gamma , p^i}$, where $i < j.$ Then $c= p^i c'$ for some $c'$, where $(c',p) = 1.$ Also assume $i> \beta -\alpha$ and $\beta +k-j \ge \alpha .$

    For any $y = p^{\beta }k' \in p^{\beta }{\mathbb {Z}},$

    $$\begin{aligned} y (u,v,w)= & {} (0,0,p^{\beta -i}p^ic')k'wc'^{-1}\\= & {} (p^{\beta -i}p^ka',p^{\beta -i}p^jb',p^{\beta -i}p^i c')k'wc'^{-1} \in {\mathbb {Z}}(a,b,c) \end{aligned}$$

    as $\beta -\alpha < i.$ So, $p^{\beta }{\mathbb {Z}} \subset [(a,b,c): G].$

    Now by considering $y \in [(a,b,c):G]$, we have $y (0 ,1, 0)= n (p^ka',p^jb',p^ic')$ for some integer n. This implies $p^{\gamma -i}\mid n$ and $y = p^{j}nb' (\bmod p^{\gamma } )$. Since $i < \gamma - \beta +j,$ so $y\in p^{\beta } {\mathbb {Z}}.$

  • Let $c \in {\mathcal {O}}_{\gamma , p^i}$, where $i < j.$ Then $c= p^i c'$ for some $c'$ such that $(c',p) = 1.$ Also assume $\beta -\alpha +k>i> \beta -\alpha$ and $j> \beta -\alpha +k.$ For any $y = p^{\beta }k' \in p^{\beta }{\mathbb {Z}},$ then

    $$\begin{aligned} y (u,v,w)= & {} (0,0,p^{\beta -i}p^ic')k'wc'^{-1}\\= & {} (p^{\beta -i}p^ka',p^{\beta -i}p^jb',p^{\beta -i}p^i c')k'wc'^{-1} \in {\mathbb {Z}}(a,b,c) \end{aligned}$$

    as $\beta -\alpha +k> i.$ So $p^{\beta }{\mathbb {Z}} \subset [(a,b,c): G].$

    Now by considering $y \in [(a,b,c):G]$, we have $y (0 ,1, 0)= n (p^ka',p^jb',p^ic')$ for some integer n. This implies $p^{\alpha -k}\mid n$ and $y = p^{j}nc' (\bmod p^{\beta })$. Since $j > \beta -\alpha +k,$ so $y\in p^{\beta } {\mathbb {Z}}.$

  • Let $c \in {\mathcal {O}}_{\gamma , p^i}$, where $i < j.$ Then $c= p^i c'$ for some $c'$, where $(c',p) = 1.$ Also assume $\gamma -\alpha +k> i \ge \beta -\alpha +k$. For any $y = p^{\alpha +i-k }k' \in p^{\alpha +i-k }{\mathbb {Z}},$

    $$\begin{aligned} y (u,v,w)= & {} (0,0,p^{\alpha -k}p^ic')k'wc'^{-1}\\= & {} (p^{\alpha -k}p^ka',p^{\alpha -k}p^jb',p^{\alpha -k}p^i c')k'wc'^{-1} \in {\mathbb {Z}}(a,b,c) \end{aligned}$$

    as $\alpha -k+j> \beta .$ So, $p^{\alpha +i-k}{\mathbb {Z}} \subset [(a,b,c): G].$

    Now by considering $y \in [(a,b,c):G]$, we have $y (0 ,0, 1)= n (p^ka',p^jb',p^ic')$ for some integer n. This implies $p^{\alpha -k}\mid n$ and $y = p^{i}nc' (\bmod p^{\gamma })$. Since $i < \gamma -\alpha +k,$ so $y\in p^{\alpha -k+i} {\mathbb {Z}}.$

  • Let $c \in {\mathcal {O}}_{\gamma , p^i}$, where $i < j.$ Then $c= p^i c'$ for some $c'$, where $(c',p) = 1.$ Also assume $i \ge \gamma -\alpha +k$ and $j> \beta -\alpha +k$. By assuming that $i \ge \gamma -\alpha +k$ and follow the same proof as above, we trivially have the required result.

  • Let $c \in {\mathcal {O}}_{\gamma , p^i}$ where $\gamma - \beta +j\le i \le \gamma - 1.$ Then $c= p^i c'$ for some $c'$, where $(c',p) = 1.$ Clearly $p^{\gamma }{\mathbb {Z}} \subset [(a,b,c): G].$ Now by considering $y \in [(a,b,c):G]$, we have $y (0 ,0, 1)= n (a,b,c)$ for some integer n. This implies $p^{\beta - j}\mid n$ and since $\beta +i -j \ge \gamma$, so $y = p^inc' (\bmod p^{\gamma } )$.

    This completes the proof.

$\square$