Construct a graph having $V=\mathbb{Z}/p\mathbb{Z}$ as its set of vertices and $\{\{x,x+1\}: x \in V\}\cup \{\{x,2x\}: x \in V\}$ as its set of edges. This graph is not an expander - quite unsurprisingly, since it is induced by a solvable group of actions.

Question: what is the simplest way to show that this graph is not an expander?

An obvious strategy is to construct a set $A$ such that $|A \cup (A+1) \cup 2A| < (1+\epsilon) |A|$ (for $\epsilon$ arbitrary and $p$ large enough in terms of epsilon). How to construct a set A is less obvious.

Two possible constructions:

(a) If $p = 2^n+1$, or more generally $p = 2^n+O(1)$, then $A = $(reductions modulo $p$ of itnegers between $0$ and $p!$ with more $0$'s than $1$'s in their binary expansion) should work.

(b) For general p, J. Cilleruelo points out to me that the set $A$ constructed by Gonzalo Fiz in Proposition 3.2 of http://arxiv.org/abs/1203.2659 (based on a Lemma of Rokhlin´s) should give an answer, at least if 2 is replaced by 4 (or any other constant square).

Any other proposals? I'd like something that can be shown quickly to work in a survey or in a class.

Construct a graph having $V=\mathbb{Z}/p\mathbb{Z}$ as its set of vertices and $\{\{x,x+1\}: x \in V\}\cup \{\{x,2x\}: x \in V\}$ as its set of edges. This graph is not an expander - quite unsurprisingly, since it is induced by a solvable group of actions.

Question: what is the simplest way to show that this graph is not an expander?

An obvious strategy is to construct a set $A$ such that $|A \cup (A+1) \cup 2A| < (1+\epsilon) |A|$ (for $\epsilon$ arbitrary and $p$ large enough in terms of epsilon). How to construct a set A is less obvious.

Two possible constructions:

(a) If $p = 2^n+1$, or more generally $p = 2^n+O(1)$, then $A = $(reductions modulo $p$ of itnegers between $0$ and $p!$ with more $0$'s than $1$'s in their binary expansion) should work.

(b) For general p, J. Cilleruelo points out to me that the set $A$ constructed by Gonzalo Fiz in Proposition 3.2 of https://arxiv.org/abs/1203.2659 (based on a Lemma of Rokhlin´s) should give an answer, at least if 2 is replaced by 4 (or any other constant square).

Any other proposals? I'd like something that can be shown quickly to work in a survey or in a class.

Construct a graph having V=Z/pZ as its set of vertices and {{x,x+1}: x in V} union {{x,2x}: x in V} as its set of edges. This graph is not an expander - quite unsurprisingly, since it is induced by a solvable group of actions.

Question: what is the simplest way to show that this graph is not an expander?

An obvious strategy is to construct a set A such that |A union A+1 union 2A| < (1+epsilon) |A| (for epsilon arbitrary and p large enough in terms of epsilon). How to construct a set A is less obvious.

Two possible constructions:

(a) If p = 2^n+1, or more generally p = 2^n+O(1), then A = (reductions modulo p of itnegers between 0 and p-! with more 0s than 1s in their binary expansion) should work.

(b) For general p, J. Cilleruelo points out to me that the set A constructed by Gonzalo Fiz in Proposition 3.2 of http://arxiv.org/abs/1203.2659 (based on a Lemma of Rokhlin´s) should give an answer, at least if 2 is replaced by 4 (or any other constant square).

Any other proposals? I'd like something that can be shown quickly to work in a survey or in a class.

Construct a graph having $V=\mathbb{Z}/p\mathbb{Z}$ as its set of vertices and $\{\{x,x+1\}: x \in V\}\cup \{\{x,2x\}: x \in V\}$ as its set of edges. This graph is not an expander - quite unsurprisingly, since it is induced by a solvable group of actions.

Question: what is the simplest way to show that this graph is not an expander?

An obvious strategy is to construct a set $A$ such that $|A \cup (A+1) \cup 2A| < (1+\epsilon) |A|$ (for $\epsilon$ arbitrary and $p$ large enough in terms of epsilon). How to construct a set A is less obvious.

Two possible constructions:

(a) If $p = 2^n+1$, or more generally $p = 2^n+O(1)$, then $A = $(reductions modulo $p$ of itnegers between $0$ and $p!$ with more $0$'s than $1$'s in their binary expansion) should work.

(b) For general p, J. Cilleruelo points out to me that the set $A$ constructed by Gonzalo Fiz in Proposition 3.2 of http://arxiv.org/abs/1203.2659 (based on a Lemma of Rokhlin´s) should give an answer, at least if 2 is replaced by 4 (or any other constant square).

Any other proposals? I'd like something that can be shown quickly to work in a survey or in a class.