Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$G$ is the set of all subsets of a set $A$, under the operation of $\triangle\;$: Symmetric Difference of sets.

$A$ has at least two different elements.

I need to check if this is a group, and if it does to show if the group is abelian and/or finite.

Associativity - easy from Symmetric Difference.
Identity element - empty group.
Inverse element - each element is inverse to itself.
Am I right?

abelian?
finite?

share|improve this question
    
You are correct. –  Calvin Lin Dec 30 '12 at 16:24
    
Welcome to MSE. You might consider using more specific, helpful question titles. See this post. For example, for this question your title could be "Is the symmetric difference a group operation?" –  William DeMeo Dec 30 '12 at 16:47

3 Answers 3

up vote 6 down vote accepted

First to get clear about the set $G$: it is the set of all subsets of a set $A$. (So the elements of $G$ are sets.) And by definition, $G$ is therefore the powerset of $A$, denoted $G =\mathcal{P}(A))$. So $|G| = 2^{|A|}$, which is finite if and only if $|A|$ is finite, and is infinite otherwise.

The operation on the sets $g_1, g_2 \in G$ is the symmetric difference of $\;g_1\;$ and $\;g_2\;$ which we'll denote as $\;g_1\;\triangle\;g_2\;$ and is defined as the set of elements which are in either of the sets but not in their intersection: $\;g_1\;\triangle\;g_2 \;= \;(g_1\cup g_2) - (g_1 \cap g_2)\tag{1}$

Hence, the symmetric difference $g_i \;\triangle\; g_j \in G\;$ for all $\;g_i, g_j \in G$. ($G$ is the set of ALL subsets of $A$, so it must include any possible set resulting from symmetric difference between any arbitrary sets in $G$, which are also subsets of $A$). That is we have now established, that the symmetric difference is closed on $G$.

The power set $G$ of any set $A$ becomes an abelian group under the operation of symmetric difference:

  • Why abelian? Easy to justify, just use the definition in $(1)$ above: it's defined in a way that $g_1 \triangle g_2$ means exactly the same set as $g_2 \triangle g_1$, for any two $g \in G$.

  • As you note, the symmetric difference on $G$ is associative, which can be shown using the definition in $(1)$, by showing for any $f, g, h \in G, (f\; \triangle\; g) \triangle \;h = f\;\triangle\; (g \;\triangle\; h)$.

  • The empty set is the identity of the group (it would be good to justify this this, too), and

  • every element in this group is its own inverse. (Can you justify this, as well? Just show for any $g_i \in G, g_i\;\triangle \; g_i = \varnothing$).

The justifications for these properties is very straightforward, but good to include for a proof that the symmetric difference, together with the set $G$ as defined, form an abelian group.


So, you've covered most of the bases, but you simply want to confirm/note the closure of the symmetric difference operation on $G$ and why, add a bit of justification for the identity and inverse claims, and to address whether, or when, $G$ is finite/infinite: this last point being that the order of $G$ depends on the cardinality of $A$.

share|improve this answer
    
wow thank you so much!!! –  Anna Dec 30 '12 at 19:34
    
You're welcome, Anna! –  amWhy Dec 30 '12 at 19:48

This is intended to be a comment to amWhy's answer. However, my comment is too long to fit in the comment box, plus I suspect my comment might be of sufficient interest to some people here to warrant a more public display.

Carolyn Bean (reference below) proved a number of interesting group-theoretic results related to the symmetric difference operation, a few of which I will state here. Let $X$ be a fixed set and let ${\cal P}(X)$ be the set of all subsets of $X.$ Bean proved that $\Delta$ can be characterized as the unique group operation $*$ on ${\cal P}(X)$ such that $A*B \subseteq A \cup B$ for all $A,$ $B \in {\cal P}(X).$ Define the co-symmetric difference operation $\nabla$ on ${\cal P}(X)$ by $A \nabla B = X - (A \Delta B)$ for all $A,$ $B \in {\cal P}(X).$ Then one can show that $\left(\,{\cal P}(X),\;\nabla\right)$ is also an abelian group. Bean proved that $\nabla$ can be characterized as the unique group operation $*$ on ${\cal P}(X)$ such that $A \cap B \subseteq A*B$ for all $A,$ $B \in {\cal P}(X).$ Bean additionally proved that $\left(\,{\cal P}(X),\;\Delta,\;\cap\right)$ and $\left(\,{\cal P}(X),\;\nabla,\;\cup\right)$ are isomorphic commutative rings with identity.

Carolyn Bean, Group operations on the power set, Journal of Undergraduate Mathematics 8 #1 (March 1976), 13-17.

share|improve this answer
    
Hi Dave, that sounds interesting, is the article available in www? –  Sven Wirsing Sep 21 at 7:55
    
@Sven Wirsing: I have not previously found it on the internet, and I looked just now and didn't find it. The is not very well known, but it's available in many U.S. libraries. FYI, I first learned about the journal in the late 1970s, well before the internet. While I'm here, you might be interested in my manuscript Exotic Group Examples that I posted (don't remember when) at Math Forum. –  Dave L. Renfro Sep 22 at 15:50
    
Thanks for these exotic examples. –  Sven Wirsing Sep 22 at 17:38

You should also mention that $g_1,g_2\in G$ then so is $g_1*g_2$ but other than that it looks OK to me.

Regarding being finite - you should know that if $A$ is finite and have $n$ elements then there are $2^n$ subsets of $A$ (or give any other argument to show that your group is finite).

If $A$ is infinite then there are infinite singeltons hence your group is also infinite.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.