If I say $\{g_1,\dots,g_n\}$ freely generates a subgroup of $G$, does that mean the elements $g_i$ are all distinct?

Question

Let $G$ be a group and let $g_1,\dots,g_n$ elements of $G$. If I say that $\{g_1,\dots,g_n\}$ freely generates a free subgroup of $G$, say $H$, do I mean that the rank of $H=n$ or should I consider the case when two elements are equal (for example $g_1=g_2$) and so the rank of $H<n$ ? thank you

Answer 1

Strictly speaking, what you wrote means that the set $\{g_1,\dots,g_n\}$ freely generates $H$, so the rank of $H$ is the number of (distinct) elements of that set, which may or may not be $n$. However, people frequently abuse language and say what you wrote when they really mean that additionally the $g_i$ should all be distinct. In context, there is usually not too much risk of confusion, and as a practical matter, if you see someone write this, you should judge for yourself which interpretation would make more sense in context. I don't really know of a succinct and completely standard way to express the latter meaning; I might write it as "the elements $(g_1,\dots,g_n)$ freely generate a subgroup", writing it as a tuple to stress that you are actually saying that the map $\{1,\dots,n\}\to H$ sending $i$ to $g_i$ satisfies the universal property to make $H$ a free group on the set $\{1,\dots,n\}$. What's really going on is that a collection of (potential) "free generators of a group $H$" should not be thought of as just a subset of $H$ but as a set together with a map to $H$ (and if the group really is freely generated by a map, the map must be injective).

(This terminological difficulty is not restricted to this context. For instance, a similar issue quite frequently comes up when talking about linear independence: if $v\in V$ is a nonzero element of a vector space, then the set $\{v,v\}$ is linearly independent (since it actually has only one element), but the tuple $(v,v)$ is not.)