If G is a sequential topological group, must G x G be sequential? 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).
 A: This is consistently true, for separable Frechet $G$.
In a recent preprint, Hrusak and Ramos-Garcia (http://www.matmor.unam.mx/~michael/preprints_files/Frechet-malykhin.pdf) have produced a model of ZFC, by iterating a Laver-type forcing, where every separable Frechet group is metrizable, and in particular, the product of Frechet groups is metrizable, and thus Frechet.
I haven't read the paper, but Justin Moore pointed me to it when I asked him about this question.
A: 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).
A: This is consistently false. 
It was proved by Malyhin and Shakhmatov (see "Cartesian products of Frechet topological groups and function spaces", Acta Math. Hung. 1992) that in any model obtained by adding a Cohen real to a model of $MA + \lnot CH$, there is a sequential topological group $G$ such that $G \times G$ is not sequential. In fact they can get $G$ to be Frechet-Urysohn and hereditarily separable (which could not have been done in $ZFC$ since consistently every separable Frechet topological group is metrizable1) while $G \times G$ has uncountable tightness.
In $ZFC$ there are examples of sequential (in fact Frechet-Urysohn) topological groups $G$ and $H$ such that $G \times H$ is not sequential (in fact not countably tight). This was shown by Todorcevic in "Some applications of $S$ and $L$ combinatorics", Ann. New York Acad. Sci. 1993.
I don´t know if there is a $ZFC$ example of a sequential topological group with a non-sequential square.

1 See M. Hrušák and U.A. Ramos-García, Malykhin's problem. Adv. Math. 262 (2014), 193–212. 
