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This is a question ($Q. 6$) from the book by Louis W. Shapiro, titled: Introduction to Abstract Algebra, from sec. $1.2$.

Let $S$ be a set, and let $G$ be the set of subsets of $S$.
If $A$ and $B$ are subsets of $S$, then define $A * B = (A \cup B) - (A \cap B)$, that is, the subset of all elements in $A$ or in $B$, but not in both.

(a) Show, using Venn diagrams if you wish, that $(A * B) * C = A * (B * C)$.
(b) Show that $(G,*)$ is a group.
(c) Show that $(G,*)$ is an abelian group.
(d) If $S$ has two or three elements, how many elements will be in $G$?
(e) If $S$ has $n$ elements, how many elements will be in $G$?

If $S$ has $n$ elements, then: (e) $|G|=2^n$. So, (d) is $4, 8$ respectively for set $S$ with $2,3$ elements.

(a) If $A * B = (A \cup B) - (A \cap B)$,
so, taking l.h.s. of $(A * B) * C = A * (B * C)$, get:
$(A * B) * C = ((A \cup B) - (A \cap B))*C$
$= ((A \cup B) - (A \cap B))\cup C - ((A \cup B) - (A \cap B))\cap C$
$= (A \cup B)\cup C - (A \cap B)\cup C - (A \cup B)\cap C + (A \cap B)\cap C$
$= A \cup B\cup C - (A \cap B)\cup C - (A \cup B)\cap C + (A \cap B)\cap C$

taking r.h.s. of $(A * B) * C = A * (B * C)$, get:
$A * (B * C) = A*(B \cup C) - A*(B \cap C))$
$= A\cup(B \cup C) - A\cap(B \cup C) - A\cup(B \cap C) + A\cap(B \cap C)$
$= A \cup B\cup C - (A \cap B)\cup C - (A \cup B)\cap C + (A \cap B)\cap C$

(b) Show that $(G,*)$ is a group, so need show that $(G,*)$ satisfies the four properties:
(i) Identity $(e)$ exists, s.t. for any set $A, A*e = e*A= A\cup e - A\cap e = A$.
As there is a set $\emptyset$ as subset of any set that satisfies this property, so $e= \emptyset$.
(ii) Inverse ($C$) exists for any element (set, $A$), s.t. $A*C = A\cup C - A\cap C = e\implies A\cup C = A\cap C\implies C = A$
(iii) Closure exists, as all sets are from the $2^n$ sets, and any union or intersection will still be in the superset. Say, there are $3$ elements in $S$, so the subsets are $8$, labelled as :
x_1. $\lbrace 1,2,3\rbrace$,
x_2. $\lbrace 1,2\rbrace$,
x_3. $\lbrace 1,3\rbrace$,
x_4. $\lbrace 2,3\rbrace$,
x_5. $\lbrace 1\rbrace$,
x_6. $\lbrace 2\rbrace$,
x_7. $\lbrace 3\rbrace$,
x_8. $\emptyset$

$x_1 \cap x_2 = x_2, x_2\cap x_7 = x_8$, and so on.
But, request a more formal way to prove it.
(iv) Associativity property exists, as both commutativity (proved below in part (c)) and closure exist $\forall$ sets $A, B, C \in G \mid A*(B*C)= (A*B)*C$.
Request approach to prove (iv) by a better approach, as not possible to show failure by contradiction.


(c) To prove that $(G, *)$ is an abelian group means that $A*B$ yields the same results under swap ($B*A$).
$A*B = (A\cup B)-(A\cap B)$, while $B*A = (B\cup A)-(B\cap A)$
Due to the set union, set intersection & set difference being commutative operations; the result is the same under swap of sets for the operation.

Siong Thye Goh
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jiten
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    Your proof is fine. Note that every element has order $2$, and $G \simeq C_2 \times \cdots \times C_2$ ($n$ factors). – Nicky Hekster Apr 02 '20 at 19:14
  • @NickyHekster I hope you meant : Order of each element $=2$, as it is its own inverse wrt the given operation. But couldn't understand the second part of your statement. Request detail or answer for that. I feel you meant that the order of the group is the product of order of elements. – jiten Apr 02 '20 at 19:24
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    See here https://math.stackexchange.com/questions/17054/group-where-every-element-is-order-2 – Nicky Hekster Apr 02 '20 at 19:48
  • @NickyHekster The question in the post at: https://math.stackexchange.com/q/1646680/424260 is stated even better by the author, so as to easily link to it. But, nowhere is given a concrete example, like (as pointed to by you) here. I wish could ping you for my future questions, as to what group properties are shown by them. Am very excited to find link to group theory issues by my trivial questions. – jiten Apr 03 '20 at 01:34
  • Ah good, even better reference/post. Yes, this question about groups in which every element has order at most $2$, pops up now and then over all these years at Math StackExchange. If you have any additional questions you can always drop me a mail ([email protected]) or post your question here of course. – Nicky Hekster Apr 03 '20 at 08:40

2 Answers2

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There is still another way of proving that the symmetric difference of sets (and write $A \Delta B= (A \cup B)-(A \cap B)$) induces an abelian group structure on the power set of $S$, and that is making use of characteristic functions mod $2$.

For every subset $A \subseteq S$, $x \in S$, define the function $1_A$ as follows. $$1_A(x)=\{^{1 \text{ if } x \in A}_{0 \text{ if } x \notin A}$$ so in particular $1_{\emptyset} \equiv 0$ and $1_S \equiv 1$. It is easy to see that $1_A=1_B$ if and only if $A=B$. Also, $1_{A \Delta B}=1_A + 1_B$ mod $2$. So, to prove associativity for example: $$1_{(A \Delta B)\Delta C}=(1_A + 1_B)+1_C=1_A + (1_B+1_C)=1_{A \Delta (B\Delta C)}$$ since addition mod $2$ is associative! Hence $(A \Delta B)\Delta C=A \Delta (B\Delta C)$.

I leave the rest to you. With this you can actually easily establish an isomorphism between ($\mathcal{P}(S)$,$\Delta$) and $C_2\times \cdots \times C_2$.

Nicky Hekster
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  • In part $(a)$, what does addition of sets means?

Let $S=\{1,2,3,4,5,6,7,8\}, A=\{1,2,3,4\}, B=\{2,3,5,6\}, C=\{3,4,6,7\}$.

$A*B=\{1,4,5,6\}$

$(A*B)*C=\{1,3,5,7\}.$

Now let's check your claim that $(A*B)*C=A \cup B \cup C -(A \cap B) \cup C-(A\cup B) \cap C +(A \cap B) \cap C$, whatever $+$ means, also assuming that $\cap$ and $\cup$ has a higher precedence than $-$.

$A\cup B\cup C = \{1,2,3,4,5,6,7\}$

$(A \cap B) \cup C = \{2,3,4,6,7\}$ and $(A \cup B) \cap C=\{3,4,6\}$ and $(A \cap B) \cap C=\{3\}$

$A \cup B \cup C - (A \cap B) \cup C = \{1,5\}$

$A \cup B \cup C - (A \cap B) \cup C - (A \cup B) \cap C = \{1,5\}$

$A \cup B \cup C - (A \cap B) \cup C - (A \cup B) \cap C + (A\cap B) \cap C = \{1,5\}+\{3\}$

which is unlikely to be correct.

  • Also, in your working, you assume that you have $(A-B) \cup C = (A \cup C)-B \cup C$, the assumption is flawed.

  • Closure: For any $A, B \subset S$, we have $A \cup B \subset S$, $A \cap B \subset S, A^c \subset S$, hence $A*B =(A \cup B) - (A \cap B)=(A \cup B) \cap (A \cap B)^c \subset S$

  • The goal of part (a) is to prove associativity.

  • Associativity: We know that $A*B = (A \cap B^c) \cup (A^c \cap B)$

Hence \begin{align}(A*B)*C &= ((A \cap B^c) \cup (A^c \cap B))*C\\ &=(((A \cap B^c) \cup (A^c \cap B)) \cap C^c ) \cup (((A \cap B^c) \cup (A^c \cap B))^c \cap C)\\ &=(A \cap B^c \cap C^c) \cup (A^c \cap B \cap C^c) \cup (((A^c \cup B) \cap (A \cup B^c)) \cap C)\\ &= (A \cap B^c \cap C^c) \cup (A^c \cap B \cap C^c) \cup ((A^c \cap B^c) \cup (A \cap B)) \cap C)\\ &= (A \cap B^c \cap C^c) \cup (A^c \cap B \cap C^c) \cup (A^c \cap B^c \cap C) \cup (A \cap B \cap C) \\\end{align}

Also, by commutivity that you have shown, \begin{align}A*(B*C)&=(B*C) *A \\ &=(C * B) *A \\ &= (C \cap B^c \cap A^c) \cup (C^c \cap B \cap A^c) \cup (C^c \cap B^c \cap A) \cup (A \cap B \cap C)\end{align}

the two expressions are equal, hence it is associative.

Edit: An alternative:

\begin{align}(A * B) * C &= ((A \cup B) - (A \cap B))*C \\&= ((A \cup B) - (A \cap B))\cup C - ((A \cup B) - (A \cap B))\cap C \\&= (((A\cup B) \cap (A \cap B)^c ) \cup C ) \cap (((A\cup B) \cap (A \cap B)^c )\cap C)^c \\&= (((A\cup B) \cap (A \cap B)^c ) \cup C ) \cap (((A \cap B^c) \cup (A^c \cap B) )\cap C)^c\\ &=(((A\cup B) \cap (A \cap B)^c ) \cup C ) \cap (((A^c \cup B) \cap (A \cup B^c) )\cup C^c)\\ &=(A\cup B \cup C) \cap (A^c \cup B^c \cup C ) \cap (A^c \cup B \cup C^c) \cap (A \cup B^c \cup C^c) \\ \\\end{align}

Again, by commutative,

\begin{align} (A*B)*C &= C*(B*A) \\ &=(A\cup B \cup C) \cap (C^c \cup B^c \cup A ) \cap (C^c \cup B \cup A^c) \cap (C \cup B^c \cup A^c) \end{align}

Siong Thye Goh
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