**Theorem.** Let $G$ be a group with a normal subgroup $H$. A positive integer $k$ and a sequence $\left(a_i\right)_{i=1}^k$ of elements of $G$ are given. A *pair product* of the $a_i$'s is an element of the form $a_ia_j$ for $i,j\in\{1,2,\ldots,k\}$ with $i\neq j$. Denote by $\ell$ the cardinality of the set of elements of order $2$ of the factor group $G/H$, and let $p$ and $q$ be positive integers.

**(a)** If one of the following conditions holds:

- $p\geq 3$, $q\geq 2$, $G/H$ has an element of order not equal to $1$ or $2$, and $$k\geq\min\big\{(p-3)\ell+(p+2q-4),pq-p-q+2\big\}\,,\tag{1}$$
- $q\geq 2$ and all non-identity elements of $G/H$ are of order $2$ and $$k\geq\min\big\{(p-1)\ell+p,pq-p-q+2\big\}\,,\tag{2}$$

then there exists a subsequence of $\left(a_i\right)_{i=1}^k$ of length $p$ with all pair products in $H$ or a subsequence of length $q$ with all pair products not in $H$.

**(b)** If one of the following conditions holds:

- $p\geq 3$, $q\geq 2$, $a_1,a_2,\ldots,a_k\notin G\setminus H$, and $$k\geq \min\big\{(p-3)\ell+(2q-1),pq-p-q+2\big\}\,,\tag{3}$$
- all non-identity elements of $G/H$ are of order $2$ and $$k\geq\min\big\{(p-1)\ell+1,pq-p-q+2\big\}\,,\tag{4}$$

then there exists a subsequence of $\left(a_i\right)_{i=1}^k$ of length $p$ with all pair products in $H$ or a subsequence of length $q$ with all pair products not in $H$.

**Sharpness.** The bounds above are sharp. If $\ell\geq q-1$, then take $g_1,g_2,\ldots,g_{q-1}\notin H$ such that the elements $g_iH\in G/H$ are pairwise distinct and of order $2$. Let $$a_{(p-1)\mu+\nu+1}:=g_{\mu+1}$$ for each $\mu=0,1,2,\ldots,q-2$ and $\nu=0,1,2,\ldots,p-2$. Then, it is easy to see that the sequence $\left(a_1,a_2,\ldots,a_{pq-p-q+1}\right)$ has no desired subsequences.

From now on, we assume that $\ell<q-1$. Then, take $g_1,g_2,\ldots,g_\ell\notin H$ such that the elements $g_iH\in G/H$ are pairwise distinct and of order $2$. If $a_1,a_2,\ldots,a_k\notin H$ is required, then we can take
$$a_{(p-1)\mu+\nu+1}:=g_{\mu+1}$$ for all $\mu=0,1,2,\ldots,\ell-1$ and $\nu=0,1,\ldots,p-2$. This shows that (4) is sharp. If $G/H$ has an element $xH$ with $x\in G$ such that $xH$ is non-identity and not of order $2$ in $G/H$, then we can also further add
$$a_{(p-1)\ell+2j-1}:=x\text{ and }a_{(p-1)\ell+2j}:=x^{-1}$$
for $j=1,2,\ldots,q-1-\ell$. Ergo, (3) is sharp.

Now, we allow $a_1,a_2,\ldots,a_k$ to be in $H$. Then, we let $a_1,a_2,\ldots,a_{(p-1)\ell}$ be as in the paragraph above. If $G/H$ has no elements of order not equal to $1$ or $2$, then set
$$a_{(p-1)\ell+j}:=1_G$$
for $j=1,2,\ldots,p-1$. Thus, (2) is sharp. If $G/H$ has an element $xH$ of order not equal to $1$ or $2$, then we further take
$$a_{(p-1)(\ell+1)+2j-1}:=x\text{ and }a_{(p-1)(\ell+1)+2j}:=x^{-1}$$
for $j=1,2,\ldots,q-2-\ell$. Thence, (1) is sharp.

**Trivial Cases.** If $p=1$ or $q=1$, then we have the obvious bound $k\geq 1$. If $p=2$ and $q\geq 2$, then (1) should be replaced by $k\geq q+1$ and (2) by $k\geq \min\{\ell+2,q\}$, whereas (3) should be replaced by $k\geq q$ and (4) by $k\geq\min\{\ell+1,q\}$. These trivial bounds are clearly sharp.

**Proof of the Theorem.** Here, we assume that $p\geq 3$ and $q\geq 2$. Let $$u:=\Big|\big\{i\in\{1,2,\ldots,k\}\,|\,a_i\in H\big\}\Big|\,,$$ $$v:=\Big|\big\{i\in\{1,2,\ldots,k\}\,|\,a_i\notin H\text{ and }a_i^2\in H\big\}\Big|\,,$$ and $$w:=k-u-v=\Big|\big\{i\in\{1,2,\ldots,k\}\,|\,a_i^2\notin H\big\}\Big|\,.$$ Suppose that the sequence $a_1,a_2,\ldots,a_k$ has neither a subsequencs of length $p$ with pair products in $H$ nor a subsequence of length $q$ with pair products not in $H$. The proof is done via contrapositivity.

We shall first prove the claim when $a_1,a_2,\ldots,a_k\notin H$ (i.e., when $u=0$). Then, it is clear that $$\left\lceil\frac{v}{p-1}\right\rceil+\left\lceil\frac{w}{2}\right\rceil\leq q-1\text{ and }\left\lceil\frac{v}{p-1}\right\rceil\leq \min\{\ell,q-1\}\,.$$ Consequently, if $t:=\left\lceil\frac{v}{p-1}\right\rceil$, then $w\leq 2(q-1-t)$ and $t\leq \min\{\ell,q-1\}$, so that, if $p\geq 3$, we have $$\begin{align}k&=v+w\leq (p-1)t+2(q-1-t)=(p-3)t+2q-2\\&<\min\big\{(p-3)\ell+(2q-1),pq-p-q+2\big\}\\&\leq \min\big\{(p-3)\ell+(p+2q-4),pq-p-q+2\big\}\,.\end{align}$$ If $G/H$ has no non-identity element of order not equal to $2$, then $w=0$, whence $$\begin{align}k&=v\leq (p-1)t\\&<\min\big\{(p-1)\ell+1,pq-p-q+2\big\}\\&\leq \min\big\{(p-1)\ell+p,pq-p-q+2\big\}\,.\end{align}$$

From now on, we assume that $u> 0$. Then, it is obvious that $u\leq p-1$. Also, we have $$\left\lceil\frac{v}{p-1}\right\rceil+\left\lceil\frac{w}{2}\right\rceil\leq q-2\text{ and }\left\lceil\frac{v}{p-1}\right\rceil\leq \min\{\ell,q-2\}\,,$$ provided that $q\geq 2$. Let $t$ be as defined before. Then, $$\begin{align}k&=u+v+w\leq (p-1)+(p-1)t+2(q-2-t)=(p-3)t+(p+2q-5)\\&<\min\big\{(p-3)\ell+(p+2q-4),pq-p-q+2\big\}\,.\end{align}$$ Finally, if all non-identity elements of $G$ have order $2$, then $w=0$ and $$k=u+v\leq (p-1)+(p-1)t<\min\big\{(p-1)\ell+p,pq-p-q+2\big\}\,.$$

In particular, let $G$ be a group with a proper normal subgroup $H$ such that, for all $g\in G$, $g^2\in H$ implies $g\in H$. For any integer $n\geq 3$ and for a sequence $\left(a_i\right)_{i=1}^k$ of elements in $G$, if $$k\geq 3n-4\,,$$ then either there exists a subsequence of length $n$ such that the product of any two entries is in $H$, or a subsequence of length $n$ such that the product of any two entries is not in $H$. Furthermore, if $a_1,a_2,\ldots,a_k\in G\setminus H$ and $$k\geq 2n-1\,,$$ then there exists a subsequence of length $n$ such that the product of any two terms is not in $H$. Both bounds are sharp.

**P.S.** For a fixed integer $d\geq 3$, what would happen if pair products are replaced by *$d$-ary products* $a_{i_1}a_{i_2}\ldots a_{i_d}$ where $i_1,i_2,\ldots,i_d\in\{1,2,\ldots,k\}$ are pairwise distinct?