We are interested in a sum-product type estimate. Let $p$ be an odd prime, and let $A$ be the order $p-1$ subgroup of $(\mathbb{Z}/p^2\mathbb{Z})^\times$. That is, let $A = \langle g^p \rangle$, where $(\mathbb{Z}/p^2\mathbb{Z})^\times = \langle g \rangle$. We seek an estimate of the size of the sum set
$$|A + A|.$$
The standard references, e.g. *Additive Combinatorics* by Tao & Vu, do not discuss this problem in the setting of $\mathbb{Z}/p^2\mathbb{Z}$, but instead focus on analogous results in finite fields. Similar estimates address situations where $|A + A|$ is small (e.g., on the order of $K|A|$), whereas in our situation, the product set $|A\cdot A|$ is very small, and we wish to show that $|A+A|$ must be large. Do results exist that address this problem more directly?



A sum-product estimate in Z/p^2Z
Heath-Brown and Konyagin stop just short of proving that $|A+A|\gg |A|^{3/2}$ in [their paper](http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.155.3891) on Gauss sums and Heilbronn's sum.


(Here $A$ is the same subgroup as above; since $A\cup\{0\}=\{0^p,\ldots,(p-1)^p\}$, we may write Heilbronn's sum as $H\_p(a)=\sum\_{n=0}^{p-1}e(an^p/p^2)=1+\sum\_{x\in A}e(ax/p^2)$.)


By Cauchy-Schwarz, we have $|A\pm A|\geq |A|^4/E\_+(A)$, where $E\_+(A)$ is the cardinality of the set of "additive quadruples" $\{(a,b,c,d)\in A^4\colon a+b=c+d\}$.


Using Stepanov's method, Heath-Brown and Konyagin show that $E\_+(A)\ll (p-1)^2+(p-1)p^{3/2}$. (This follows from their lemma 4 and the argument at the bottom of page 7.)


Combining the previous two results yields $|A\pm A|\gg |A|^{3/2}$, where the hidden constant is $1+o(1)$. In [http://arxiv.org/abs/math/0304217], Konyagin proves the same bound for multiplicative subgroups of $\mathbb{F}\_p$ (see lemma 5).


Shkredov gives the improved bound $E\_+(A)\ll |A|^{42/17}(\log|A|)^{10/17}$ in [http://arxiv.org/abs/1208.6124v1], and a further improvement $E\_+(A)\ll |A|^{22/9}(\log|A|)^{2/3}$ in [[S]](http://arxiv.org/abs/1302.3839). Thus up to logs, our additive energy argument gives $|A\pm A|\gg |A|^{3/2+1/18}$.


You can do better using an argument of Vyugin and Shkredov [VS] from [http://arxiv.org/abs/1102.1172]. Theorem 5.5 from [VS] shows that $|R\pm R|\gg |R|^{5/3}(\log|R|)^{-1/2}$, where $R$ is a multiplicative subgroup of $\mathbb{F}\_p$ of size at most $p^{1/2}$. The proof relies on the following results from [VS]: Lemma 2.3 and Corollary 2.7, which are true for any finite abelian group, and Corollary 5.1 and Lemma 5.4, whose analogs in [S] are Proposition 8 and Corollary 9. Thus we have $|A\pm A|\gg |A|^{5/3}(\log|A|)^{-1/2}$.


Finally, you can apply Proposition 8 from [S] to get a lower bound on |A+A+A|.



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> Proposition 8 ([S]) Let $A$ be the multiplicative subgroup of $\mathbb{Z}\_{p^2}$ of order $p-1$. Suppose that $Q\_1,Q\_2,Q\_3$ are $A$-invariant subsets of $\mathbb{Z}\_{p^2}$ and $|Q\_1||Q\_2||Q\_3|\ll p^5$. Then
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> > > $\sum\_{z\in Q\_1}Q\_2\ast (-Q\_3)(z)\ll p^{-1/3}(|Q\_1||Q\_2||Q\_3|)^{2/3}.$
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(Note that Shkredov uses $\circ$ to denote convolution with $-B$; we're also using the convention that $B(x)$ is the indicator function of $B$.)


If $Q\_1=Q\_2-Q\_3$, then the left hand side of the above equation is equal to $|Q\_2||Q\_3|$. Choosing $Q\_1=A+A+A$, $Q\_2=A$ and $Q\_3=-(A+A)$, we get that



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> > $|A+A+A|\gg p^{1/2}|A|^{1/2}|A+A|^{1/2}\gg p^{11/6}/(\log|A|)^{1/4}$. 
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