Does the Symmetric difference operator define a group on the powerset of a set?

$G$ 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?

Solution 1:

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 $A$.

Solution 2:

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.

Solution 3:

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.