TITLE: If G is a sequential topological group, must G x G be sequential?
QUESTION [13 upvotes]: Using standard definitions, the topological space $Y$ is sequential if for each nonclosed $A \subset Y$, there exists a convergent sequence $a_{1}$ , $a_{2}$,...$\rightarrow b$
so that $a_{n} \in A$ but $b \notin A$.  
Working in the topological category TOP, we assume the group $G$ is a sequential space, inversion is continuous, and group multiplication is continuous with respect to the standard product topology on $G \times  G$. 
Must $G \times G$ be a sequential space?
A `yes' answer would provide a sharp dividing line between sequential topological groups in TOP, and sequential topological groups in SEQ\TOP. 
(SEQ is a category in which the standard product topology of $G \times G$ is refined to ensure $G \times G$ is sequential).

REPLY [2 votes]: A simple example of sequential topological groups $G,H$ with non-sequential product $G\times H$: $G=\mathbb R^\omega$ and $H=\mathbb R^\infty=\lim \mathbb R^n$ be the direct limit of finite-dimensional Euclidean spaces. The group $G$ is metrizable and $H$ is a sequential $k_\omega$-space. The non-sequentiality of $G\times H$ follows from a result [Banakh, Taras; Zdomskyĭ, Lubomyr. The topological structure of (homogeneous) spaces and groups with countable $cs^*$-character. Appl. Gen. Topol. 5 (2004), no. 1, 25--48; MR2087279] saying that each sequential topological group $G$ of countable $cs^*$-character is either metrizable or contains an open $k_\omega$-subgroup.
We say that a topological space $G$ has countable $cs^*$-character if for every point $x\in X$ there is a countable family $\mathcal N$ of subsets of $X$ such that for every neighborhood $O_x\subset X$ and every sequence $(x_n)$ convergent to $x$ in $X$ there is a set $N\subset O$ in the family $\mathcal N$ containing infinitely many points of the sequence $(x_n)$.
The mentioned result of Banakh and Zdomskyy implies that for any sequential topological group $G$ of countable $cs^*$-character the square $G\times G$ is sequential (more precisely, $G\times G$ is metrizable or contains an open $k_\omega$-subgroup).