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).



If G is a sequential topological group, must G x G be sequential?
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.




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


