10
$\begingroup$

What is the difference between a Subgroup and a subset? I know hardly any Abstract algebra, just some things from youtube and wikipedia, but the notion of a subgroup being part of a larger group and a set being part of some group is indistinguishable to me. A nice simple answer would do fine. Thank you for your time

$\endgroup$
4
  • 3
    $\begingroup$ It seems to me that you think that a group and a set are the same thing, hence your confusion regarding subgroups and subsets. Sets and groups are different concepts. Are you aware of this? $\endgroup$
    – Git Gud
    Commented Jan 12, 2013 at 23:26
  • $\begingroup$ I'm not sure if this is correct but, Groups have mathematical properties associated with them and the sets are the elements inside of the group? Then is it correct to say $\mathbb{R}^2$ is a subgroup of $\mathbb{R}^3$ $\endgroup$ Commented Jan 12, 2013 at 23:28
  • $\begingroup$ Nah, man. Glad this is clear now. You're confused about somethings. I'm not really sure on how to explain this to you. Groups and sets are entirely different things. Sets are the core of anything mathematical. I'm sure Brian will read your comment and he'll know of a way to explain this to you in a way you can understand. $\endgroup$
    – Git Gud
    Commented Jan 12, 2013 at 23:33
  • $\begingroup$ Anyway, depending on how serious you are about learning mathematics, you should take a look at one of these books: amazon.com/How-Prove-It-Structured-Approach/dp/0521446635 amazon.com/Elements-Advanced-Mathematics-Edition-Studies/dp/… $\endgroup$
    – Git Gud
    Commented Jan 12, 2013 at 23:49

4 Answers 4

17
$\begingroup$

Consider the group $\Bbb Z$ of integers under addition. The subset $S=\{1,2\}$ is not a subgroup. It isn’t a group at all: it isn’t closed under the operation (addition), since $2+2$ is not in $S$, and it doesn’t have an additive identity.

The set $E$ of all even integers, on the other hand, is a subset of $\Bbb Z$ that is a subgroup: it’s a group in its own right using the same operation, addition, as $\Bbb Z$.

If $G$ is any group, every subgroup of $G$ is by definition a subset of $G$, but as the example above shows, not every subset of $G$ need be a subgroup.

$\endgroup$
1
  • $\begingroup$ +1. It might be worth noting, though, that this is slightly informal: a group is formally defined as a structure consisting of a set and a group operation, such that formally, a group is not exactly the same as the set on which it's defined. So while it's fine, normal, and common to say things like "every subgroup is a subset", that's not 100% formally precise. $\endgroup$
    – ruakh
    Commented Jan 13, 2013 at 5:00
9
$\begingroup$

Groups are sets with additional structure: a binary operation which has certain properties.

A subgroup is a subset which is also a group of its own, in a way compatible with the original group structure. But not every subset is a subgroup. To be a subgroup you need to contain the neutral element, and be closed under the binary operation, and the existence of an inverse.

For example, groups are never empty (they have a neutral element), so the empty set is always a subset but never a subgroup. The rational numbers are a subgroup of the real numbers, and a subset of the real numbers, whereas $\{0,1\}$ is a subset but not a subgroup, $1+1\neq 0$. Similarly, $\mathbb N$ is a subset of $\mathbb R$ which is not a group, despite being closed under addition, $1$ does not have an additive inverse (which is $-1$) in $\mathbb N$.

$\endgroup$
2
  • $\begingroup$ and closed under inversion, maybe? $\endgroup$
    – RiaD
    Commented Jan 13, 2013 at 6:53
  • $\begingroup$ @RiaD: Of course, but I was trying to remain a bit vague for a reason. If one is trying to be precise then one has to give all the definitions. I'll how I can edit to add your (important) point. Thank you! $\endgroup$
    – Asaf Karagila
    Commented Jan 13, 2013 at 6:54
7
$\begingroup$

You have laws that make a set $S$ become a group. You must have an operation $\star$ associated with this set which is closed under this operation (if $x,y\in S$ then $x \star y \in S$) and the operation must have these three properties :

  1. If $a,b,c\in S$ then $a\star(b\star c)=(a\star b)\star c$
  2. There exists a (unique) identity $e_S$ such that $a\star e_S=e_S\star a=a$ for every $a\in S$
  3. For every $a\in S$, there exists a (unique) inverse $b$ such that $a\star b=b\star a=e_S$

If all these conditions are filled, then $S$ is called a group.

A subset of $S$ is just a subset of $S$ as a set.

But a subgroup of $S$ must have the three properties above.

$\endgroup$
1
$\begingroup$

A subgroup $H$ of $G$ it's only a non-empty subset of $G$ such that $x - y \in H$ if $x,y \in H$, that's to say, a subset of $G$ with additional properties. For example, by definition the empty set is not a subgroup.

$\endgroup$
4
  • $\begingroup$ That is true if you use the notation of addition and subtraction, but generally that's used only when the group is abelian. $\endgroup$ Commented Jan 13, 2013 at 0:07
  • $\begingroup$ Well, it's only a notation. $\endgroup$ Commented Jan 13, 2013 at 0:09
  • 3
    $\begingroup$ Another issue I have with this is that although it's an efficient way of saying it, maybe the less efficient way is the easier way to understand: i.e. it's closed both under the group operation and under inversion. $\endgroup$ Commented Jan 13, 2013 at 0:19
  • $\begingroup$ You are right, but maybe it's a good exercise for a beginner in algebra. $\endgroup$ Commented Jan 13, 2013 at 0:29

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .