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$G$ 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?

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 $G =mathcalP(A))$. So $|G| = 2^$, 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: $;g_1;triangle;g_2 ;= ;(g_1cup g_2) - (g_1 cap g_2)tag1$

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 $A$.

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 $left(,cal P(X),;nablaright)$ 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 $left(,cal P(X),;Delta,;capright)$ and $left(,cal P(X),;nabla,;cupright)$ are isomorphic commutative rings with identity.

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

You should also mention that $g_1,g_2in G$ 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.