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

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

• If the elements freely generate $\;H\;$ it can't be $\;g_i=g_j\;,\;\;i\neq j$, nor $\;g_i=1\;$ for some $\;i\;$ – DonAntonio Feb 28 '16 at 1:31
• @Joanpemo I agree that it can't be $g_i=1$ but why it can't be $g_i=g_j$ for $i\not=j$? thank you – Richard Feb 28 '16 at 1:34
• Else you would have the ralation $g_ig_j^{-1}$. – Daniel Robert-Nicoud Feb 28 '16 at 1:38
• @Joanpemo It depends on the context, but the way the question is phrased, i.e., without context, the statement is not strictly false. There is a perfectly nice free group which is free on the set $\{x,x,y\}$. It is the free group on $\{x,y\}$, since $\{x,x,y\}=\{x,y\}$. – Zach Blumenstein Feb 28 '16 at 1:44
• @ZachBlumenstein Thank you. Yet, as sets, $\;\{x,x,y\}=\{x,y\}\;$ so no problem, indeed. I though understand something different when I read "a set of free generators", and as far as I know this implies $\;g_i\neq g_j\;$ for $\;i\neq j\;$, otherwise the universal property of a free group in the number of free generators given isn't fulfilled. – DonAntonio Feb 28 '16 at 2:42

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.)
• I don't exactly agree. When we say that a (finite or infinite) set $X$ is a set of free generators of a group $F$, we don't need to specify any function, because the function is just inclusion. There is no mistake there. But I do agree that the notation $\{g_1,\ldots,g_n\}$ taken to mean that $g_i$ are distinct is bad, because it literally means what that $X=\{g_1,\ldots,g_n\}$ is a free generating set of $F$, while it probably really means that $g_1, \ldots, g_n$ generate $F$ freely, which is not the same thing if $\lvert X\rvert\neq n$. – tomasz Feb 28 '16 at 8:34
• Sure, there is nothing wrong with saying that a subset freely generates a group. But it is more natural to formulate it in terms of maps from a set, and the failure to do so is the cause of the terminological awkwardness when saying "$(g_1,\dots,g_n)$ freely generates" a group, and the frequency with which people incorrectly write "$\{g_1,\dots,g_n\}$ freely generates" when that's not really what they mean. – Eric Wofsey Feb 28 '16 at 8:41