I know that the symmetric difference looks like that when having three sets A,B,C:    I want to make a prove with Venn diagrams to prove the associativity of the symmetric difference. What are some good ways to create this proof in a clear way?  I really appreciate your answer!
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By $A\oplus B$ we denote the symmectric difference of two sets. The definition is $A\oplus B =(A\setminus B) \cup (B\setminus A)$. Now I hope to show that $A\oplus (B\oplus C) = (A\oplus B)\oplus C$. I remember that there's an elegant proof, but I forget its detail. (The first step is to show $A\oplus (A\oplus B) = B$, and maybe applying this formula, we can obtain associativity.)
104092
Assume that $n\geq 3$. Let $S_n$ be the symmetric group with $n!$ elements. Prove that there exists $x$ and $y$ in $S_n$ such that $x$ and $y$ have order $2$ and $xy$ has order $n$.  For $n=3$, it is easy to see that $x=(1, 2)$ and $y=(1, 3)$ works. How to prove it for the general case?
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I'm trying to understand this proof and I can follow it up the the second last paragraph where it states what happens if n is odd/even. I don't understand why there is just one conjugacy class when n is odd and two when n is even.    Help much appreciated, thanks!
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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?
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Let $G$ be a finite, nonempty set with an operation $*$ such that   $G$ is closed under $*$ and $*$ is associative Given $a,b,c \in G$ with $a*b=a*c$, then $b=c$. Given $a,b,c \in G$ with $b*a=c*a$, then $b=c$.   I want to prove that $G$ is a group, but I don't know how to show that there exists an identity $e\in  G$ such that $e*x=x$ and $x*e=x$ $\forall x \in G$. I also don't know how to show that $\forall$ x $\in G$ there exists a $y \in G$ such that $y*x=e$ and $x*y=e$. How do I do this?
91065
I am trying to prove that if $A – C = B – C$ and $A\cap C = B\cap C$ then $A = B$. I have tried using Venn Diagrams as a proof technique, but we are not able to use proof by Venn Diagrams.
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How would I formally prove that for the integers $a$, $b$, and $d$, if $d\mid da+b$, then $d\mid b$? Would a direct proof be the best option? If I do a direct proof I seem to get stuck pretty quickly... in fact I don't even know how to start...   Assume $d\mid da+b$ ... ?  This might look like a HW dump but I honestly just don't know how to continue...
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Let $S$ be the relation on $\mathbb{N} \times \mathbb{N}$ defined by $(a,b)S(c,d)$ if and only if $ad=bc$. Prove this is an equivalence relation on $\mathbb{N} \times \mathbb{N}$.  I think I've found this to be reflexive and symmetric, but I'm stuck on transitivity. Can someone check my work so far and assist with testing transitivity?  Reflexive: Let $(x,y)S(x,y)$. Then $xy = yx$. So $S$ is reflexive. Symmetric: Suppose $(a,b)S(c,d)$. Then, $ad = bc$. Therefore, $da = cb$ and $cb = da$. Therefore, $(c,d)S(a,b)$. Thus, $S$ is symmetric.  Transitive: Suppose $(a,b)S(b,c)$. Then $ac = bb$.   This is where I'm stuck. Any ideas?
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Give short arguments for each part below to complete the reasoning. Please do not use Venn diagrams.     a. Prove that $A = (A\setminus B) \cup (A \cap B)$.   (Hint: You should prove that $(A\setminus B)\cup (A\cap B) \subseteq A$ and vice-versa)      b. Prove that $(A \setminus B) \cap (A \cap B) = \emptyset$   (Hint: Try a proof by contradiction for this statement.).    I'm not too sure exactly where to start on this. Both statements make sense logically, but I don't know how exactly to prove them. Can anyone point me in the right direction?
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I need help with a proof for my liner algebra class.     If $A$ is a square matrix, then $AA^T$ and $A^TA$ are symmetric.   I have no idea where to start!
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I'm trying to figure out a combinatorial proof for:  $$\displaystyle \sum_{k \mathop = 0}^n \binom {r+k} k = \binom {r + n + 1} n$$  I've tried the committee counting thing, but that didn't work.
55171
Show that if $a, b$ and $c$ are integers with $c|ab$ then $c|(a,c)(b,c)$.  Now $(a, c)$ and $(b, c)$ would both divide $c$ since it's the gcd, but how would I show $c$ divides their product, and $(a,c)(b, c) \geq  c$ and not the other way around?
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I have to resit a calculus exam and for some reason set proofs were never my best friend...  Anyway, on a practice exam I encountered the following proof:  $$A\cap(B\cup C) = (A\cap B)\cup(A\cap C)$$  When I draw a Venn-diagram it seems quite obvious but I couldn't manage to write the proof down properly.  If someone could help me, that'd be great!
47772
Prove that, for $n \geq 3$, the three-cycles generate the alternation group $A_n$   Proof:  We multiply on the left by 3-cycles to "reduce" an even permutation $p$ to the identity, using induction on the number of indices fixed by a permutation. How the indices are numbered is irrelevant. If $p$ contains a $k$-cycle with $k \geq 3$, we may assume that it has the form $p=(123\dots k)\dots$ Multiplying on the left by $(321)$ gives $$p'= (321)(123 \dots k)\dots=(1)(2)(3\dots k)\dots$$ More fixed indices.   What do you think ?
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$A$$\rightarrow$$(B$ $\vee$ $C$ ) , $B$ $\rightarrow$ $C$ $\vDash$ $A$ $\rightarrow$ $D$  I think it's wrong but I have no idea how to prove.
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I'd like to know a reference for a simple proof that $\{c_\mu\mid \mu\vdash n\}$ is a basis for the centre of the symmetric group algebra $\mathbb{C}\mathfrak{S}_n$, where $c_\mu$ is the sum of all elements of a given cycle type (i.e. the sum of every element in the conjugacy class corresponding to $\mu$).
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Related Link:    Reference: Fraleigh p. 49 Question 4.38 in A First Course in Abstract Algebra     I will present my proof (distinct from those in the link) for critique and then ask my question.  $G$ is a set and $ \times $ is an associative binary operation.  Suppose that there exists a $e \in G$ such that, for all $a \in G$, $ea = a$ and $a^{-1}a = e$ for some $a^{-1} \in G$.  Show that for the same $e$, $ae = a$ and $aa^{-1} = e$.  $a^{-1}(aa^{-1})a = (a^{-1}a)(a^{-1}a) = ee = e = a^{-1}a = a^{-1}(a^{-1}a)a$  Since $a^{-1} \in G$, it has a left inverse; apply it to both ends, and we have $(aa^{-1})a = (a^{-1}a)a$.  As a result, $ae$ = $a(a^{-1}a) = (aa^{-1})a = (a^{-1}a)a = ea$.  For the right inverse, start with $aa^{-1} = a(a^{-1}a)a^{-1} = (aa^{-1})(aa^{-1})$.  Since $\times$ is a binary operation, $aa^{-1} \in G$ and has a left inverse; apply it to both ends, and we have $e = aa^{-1}$.  In second comment following the question in the link, Mr. Derek Holt pointed out that the requester did not word his/her question correctly.  Specifically, the identity in the second axiom is not well-defined.     Let $(G, *)$ be a semi-group. Suppose    1. $ \exists e \in G$ such that $\forall a \in G,\  ae = a$;    2. $\forall a \in G, \exists a^{-1} \in G$ such that $aa^{-1} = e$.    How can we prove that $(G,*)$ is a group?    This formulation makes the same technical error as many textbooks. The $e$ in your second axiom is not well-defined. "But obviously it's intended to be the same $e$ as in the first axiom" you reply. But the first axiom does not necessarily specify a unique element $e$. So should we interpret the second axiom as meaning "for some $e$ as in 1" or "for all $e$ as in 1"? –  Derek Holt Sep 17 '11 at 15:31   Was he saying that if, in axiom 1, we have $ae_1 = a, ae_2 = a$, but $e_1 \neq e_2$,  when we get to axiom 2, do we have $aa^{-1} = e_1, aa^{-1} = e_2$, or two different inverses so that $aa_1^{-1} =  e_1, aa^{-1}_2 = e_2$?  I think my wording eliminated the ambiguity.  It does not imply that $e$ is unique, but if $e$ is a left identity and produces left inverses, then it is also a right identity and produces right inverses. I tried really hard on this one; please kindly point out my mistakes.