Let's consider the set consisting of $8$ elements $\{\pm1, \pm i,\pm j, \pm k\}$ with the following multiplication table:     We see that $1$ commutes with any of the $\{i,j,k\}$.  For example, we want to consider the product $(-1)\cdot j$, since $i^2=-1$ then we can rewrite it as: $(-1)\cdot j=(i\cdot i)\cdot j=i\cdot(i\cdot j)=i\cdot k=-j$.   In this example and in many others in order to multiply elements we need to use associativity property. How to prove that associativity is true in this set?  In similar  I have seen approach using automorphism. However, I was not able to comprehend it. Can anyone explain it please?   Would be grateful if somebody can demonstrate some elementary approach. In my opinion it is definitely important to know.
61521
Consider the eight objects $\pm 1, \pm i, \pm j, \pm k$ with multiplication rules:   $ij=k,jk=i,ki=j,ji=-k,kj=-i,ik=-j,i^2=j^2=k^2=-1$,   where the minus signs behave as expected and $1$ and $-1$ multiply as expected. Show that these objects form a group containing exactly one involution.  Well, it is easy to determine closure from the definition. $1$ is clearly the identity, and the inverses can also be determined $(i,-i),(j,-j),(k,-k)$, $-1$ with itself, and $1$ with itself. The only involution is $-1$. Is there an easy way to check associativity of this group? There are too many possible combinations $(ab)c=a(bc)$ to check directly. ($8^3$ possible combinations)
73537
This is one of my first proofs about fields. Please feed back and criticise in every way (including style and details).  Let $(F, +, \cdot)$ be a field. Non-trivially, $\textit{associativity}$ implies that any parentheses are meaningless. Therefore, we will not use parentheses. Therefore, we will not use $\textit{associativity}$ explicitly.  By $\textit{identity element}$, $F \ne \emptyset$. Now, let $a \in F$. It remains to prove that $0a = 0$. \begin{equation*} \begin{split} 0a &amp;=  0a +  0      &amp;&amp; \quad \text{by }\textit{identity element }(+    ) \\    &amp;=  0a +  a + -a &amp;&amp; \quad \text{by }\textit{inverse element  }(+    ) \\    &amp;=  0a + 1a + -a &amp;&amp; \quad \text{by }\textit{identity element }(\cdot) \\    &amp;= (0 + 1)a + -a &amp;&amp; \quad \text{by }\textit{distributivity   }        \\    &amp;= (1 + 0)a + -a &amp;&amp; \quad \text{by }\textit{commutativity    }(+    ) \\    &amp;=      1 a + -a &amp;&amp; \quad \text{by }\textit{identity element }(+    ) \\    &amp;=        a + -a &amp;&amp; \quad \text{by }\textit{identity element }(\cdot) \\    &amp;=  0            &amp;&amp; \quad \text{by }\textit{inverse element  }(+    ) \end{split} \end{equation*} QED  PS: Is "Let $(F, +, \cdot)$ be a field." ok? Besides, I would not want to call $F$ a field, because $F$ is just a set. Also, what do you think about using adverbs like "Now"? How would you have said the associativity-thing?
23965
I have trouble understanding the following task:     Show that the set $X = \{f_1,\ldots ,f_6\}$ of functions $f_i : \Bbb Q\setminus \{0,1\} \to \Bbb Q\setminus \{0,1\}$ with   $x ↦ f_1(x) = x$,$x ↦ f_2(x) = 1/x$,$ x ↦ f_3(x) = 1-x,$ $x ↦ f_4(x) = 1/(1-x)$,$ x ↦ f_5(x) = (x-1)/x$, $x ↦ f_6(x) = x/(x-1)$   and the composition as the operation is a group. Is this group commutative?    I'm a little confused as I don't know what to do with all these functions. I can't apply the definitions of a group on this because of that. If the group is $(X,∘)$ how do I show that this is associative don't I need the information on how the single functions are connected or am I missing something?
77527
(Reference : Fraleigh, A first course in abstract algebra)     Prove that a nonempty set $G$, together with an associative binary operation * on $G$ such that $a*x=b$ and $y*a=b$ have solutions in G, $\forall a,b\in G$, is a group.   Because * is an associatie operation, we have to prove that: there exists $e\in G$ such that $e*a=a=a*e$ for any $a\in G$, and that for any $a\in G$ there exists $\alpha \in G$ such that $a*\alpha=e=\alpha *a$.  This is very confusing, first I tried to prove the identity exists. If we consider a=b, then $a*x=a=y*a$, and because of the hypothesis $x,y\in G$ always exists, now we have to show that $x=y$, but I just can't do this with out inverses, so I tried to prove that the inverses exists, however I have no idea how the identity looks like, I'm stuck.  And I tried to get unstuck, so I thought particular cases, first if $G$ had only one element, then what would happend if it had just two elements... but I remembered that we don't know if G is finite or not, and I think this makes things a little bit more complicated.  What can I do?
24178
This is how I'm approaching it: let $R$ be a finite integral domain and I'm trying to show every element in $R$ has an inverse:   let $R-\{0\}=\{x_1,x_2,\ldots,x_k\}$,  then as $R$ is closed under multiplication $\prod_{n=1}^k\ x_i=x_j$,  therefore by canceling $x_j$ we get $x_1x_2\cdots x_{j-1}x_{j+1}\cdots x_k=1 $,  by commuting any of these elements to the front we find an inverse for first term, e.g. for $x_m$ we have $x_m(x_1\cdots x_{m-1}x_{m+1}\cdots x_{j-1}x_{j+1}\cdots x_k)=1$, where $(x_m)^{-1}=x_1\cdots x_{m-1}x_{m+1}\cdots x_{j-1}x_{j+1}\cdots x_k$.   As far as I can see this is correct, so we have found inverses for all $x_i\in R$ apart from $x_j$ if I am right so far. How would we find $(x_{j})^{-1}$?
15984
Let $G$ be a nonempty set closed under an associative product, which in addition satisfies :      A. There exists an $e$ in $G$ such that $a \cdot e=a$ for all $a \in G$.      B. Given $a \in G$, there exists an element $y(a) \in G$ such that $a \cdot y(a) =e$.      Prove that $G$ must be a group under this product.   Attempt -Since Associativity is given and Closure also, also the right identity and right inverse is given .So i just have to prove left identity and left inverse.  Now as $ae=a$ post multiplying by a, $aea=aa$. Now pre multiply by a^{-1} I get hence $ea=a$. And doing same process for inverse Is this Right?
11969
I started thinking about the number of associative (binary) operations on a set with $n$ elements today.  Looking online I found  which indicates only $113$ of the possible $19,683$ operations on a three-element set satisfy association.  So, $50\%$ of binary operations on a two-element set satisfy association, and less than $0.6\%$ of all binary operations on a three element set satisfy association.  For an $n$-element set, where $n$ denotes a natural number, is the ratio $R_{n}=A_{n}/B_{n}$, of the number of associative binary operations $A_{n}$ to the number of all binary operations $B_{n}$, in general small?  I don't mean the following questions as equivalent, but since they seem more concretely answerable in principle, does   $$ \lim_{n \to +\infty} \frac{A_{n}}{B_{n}} = 0 ? $$       Also, is $F : n \to R_{n}$ a monotonically decreasing function?  Some Background of the Question: In his Linear Algebra Problem Book 1995 on p. 6 Paul Halmos writes "The commonly accepted attitudes toward the commutative law and the associative law are different.  Many real life operations fail to commute; the mathematical community has learned to live with that fact and even to enjoy it.  Violations of the associative law, on the other hand, are usually considered by specialists only."  For all I know, Halmos might only have written that to motivate the study of Linear Algebra and doesn't quite literally mean what he appears to say.  But, if he means what he appears to say, and if $F : n \to R_{n}$ is monotonically decreasing, or $R_{n}$ is generally small, I think there's something amiss with what Halmos says, since non-associative operations seem so common that one may as well enjoy them.
51212
Let $(A,+,\cdot)$ be a unity ring with the property that if $x \in A$ and $x^2=0$ then $x=0$. Consider the set $M=\{a\in A | a^3=a\}$. Prove that:      a) $2a\in Z(A)$, $\forall a\in M$, where $Z(A)$ denotes the centre of the ring $A$;   b) $ab=ba$, $\forall a,b\in M$.   My attempts revolved around the fact that an idempotent element in a reduced ring is central. So, since for $a\in M$ we have that $(a^2)^2=a^2$, it follows that $a^2\in Z(A)$, $\forall a\in M$. The next thing I wanted to use in order to solve a) was that $Z(A)$ is a subring of $A$, so if I had proved that $(a+1)^2 \in Z(A)$, $\forall a\in M$, then we would have reached the desired conclusion. However, I couldn't prove this and I honestly doubt that it is true. Another idea that I had was to prove that $M$ is a subring of $A$. Of course, this didn't work out because I cannot even prove that $M$ is closed under addition. Again, I don't know if this is true and it most likely isn't. As for b), I think that a) should be of use, but I don't know how. It is a well-known problem that a ring with $x^3=x$ for any $x$ in that ring is commutative, but since $(M,+,\cdot)$ is almost definitely not a ring, this doesn't help. EDIT: Is there any chance that this question is simply wrong? I tended to believe this before asking it here too, but since nobody has made any progress on it until now I am even more inclined to think so.
62086
The exercise for which I am seeking feedback wants me to prove that the center of a direct product is the direct product of the centers. That is, $$Z(G_1 \times \cdots \times G_n)=Z(G_1) \times \cdots \times Z(G_n).$$  $\textbf{Attempt at Solution:}$ We proceed by induction on $n$. First, we prove our desired for $n=2$; this will establish our basis step. Let $(g_{1},g_{2}) \in Z(G_{1} \times G_{2})$. Then $(g_{1},g_{2})(x_{1},x_{2})=(x_{1},x_{2})(g_{1},g_{2})$ for all $(x_{1},x_{2}) \in G_{1} \times G_{2}.$ This implies $g_{1}x_{1}=x_{1}g_{1}$ and $g_{2}x_{2}=x_{2}g_{2}$ for all $x_{1} \in G_{1}$ and for all $x_{2} \in G_{2}$. Hence, $g_{1} \in Z(G_{1})$ and $g_{2} \in Z(G_{2})$. To prove the reverse inclusion, this time let $(g_{1},g_{2}) \in Z(G_{1}) \times Z(G_{2})$ Then, $g_{1}x_{1}=x_{1}g_{1}$ and $g_{2}x_{2}=x_{2}g_{2}$ for all $x_{1} \in G_{1}$ and for all $x_{2} \in G_{2}$; it, of course, follows that $(g_{1},g_{2})(x_{1},x_{2})=(x_{1},x_{2})(g_{1},g_{2})$ for all $(x_{1},x_{2}) \in G_{1} \times G_{2}.$ Indeed, $(g_{1},g_{2}) \in Z(G_{1} \times G_{2})$  For our induction step, assume that desired result holds for some $n$; that is, assume $$Z(G_1 \times \cdots \times G_n)=Z(G_1) \times \cdots \times Z(G_n).$$ We show $$Z(G_1 \times \cdots \times G_n \times G_{n+1})=Z(G_1) \times \cdots \times Z(G_n) \times Z(G_{n+1}).$$ By our basis step, $$Z(G_1 \times \cdots \times G_n \times G_{n+1})=Z(G_1 \times \cdots \times G_n) \times Z(G_{n+1}).$$ Then, using our induction hypothesis, we obtain $$Z(G_1 \times \cdots \times G_n) \times Z(G_{n+1})=Z(G_1) \times \cdots \times Z(G_n) \times Z(G_{n+1}).$$ Hence, our induction step has been established. Finally, by the principle of mathematical induction, we conclude $$Z(G_1 \times \cdots \times G_n)=Z(G_1) \times \cdots \times Z(G_n),$$ for any finite collection of groups $G_{1}, \ldots , G_{n}.$  $\blacksquare$  $\textbf{Question:}$ This may be a stupid question; but, I'll ask anyway! Would it be wrong if I completed the proof by letting $(g_{1},\ldots , g_{n}) \in Z(G_1 \times \cdots \times G_n)$ and showing $(g_{1},\ldots , g_{n}) \in Z(G_1) \times \cdots \times Z(G_n)$? (Then, of course, proceed by proving the reverse inclusion.) Thank you!
26386
Let $(\mathbb K,+,\cdot)$ be a field. Is there a group isomorphism between $(\mathbb K,+)$ and $(\mathbb K^\times,\cdot) $ ?  The answer should clearly be negative. I tried to proceed via contradiction, but this has not led me very far. Since it's some kind of &quot;trick problem&quot;, I'm merely looking for hints.
54372
I am presented with the following task:     Show that if $U$ is the collection of all units in a ring $\langle R, +, \cdot\rangle$ with unity, then $\langle U, \cdot\rangle$ is a group.   I am still not confident with proofs in Abstract Algebra, and I'd like to receive some verification/slaughter/constructive critique on my reasoning here. I do not know if there are any mistakes, or how serious they are.     We know that there is an identity element, since the unity itself is a unit, and therefore in the group. It follows directly from the definition of a unit that there exists an inverse for every element in the group, and associativity follows directly from the Ring-axioms. As for closure, we need to show that the product of two units is also a unit. Let $a, b \in U$ be two elements with the property that $a\cdot b \ne e$, where $e$ is the identity element in $\langle U, \cdot \rangle$. We know that there exists elements $a', b' \in U$ with the property that $a' \cdot a = e = a' \cdot a$ and $b' \cdot b = e = b \cdot b'$. Let $a', b'$ be two such elements with the property that $a' \cdot b' \ne e$. We may now compute $(a\cdot b) \cdot (a' \cdot b')$, and from the Ring-axioms we have associativity under multiplication, so we may write the expression as $a \cdot (b \cdot a' \cdot b') = a \cdot (b \cdot b' \cdot a') = a \cdot (e \cdot a') = a \cdot a' = e$. Thus we have shown that the product of any two elements is a unity, which leads to closure. All the group-axioms are met and $\langle U, \cdot \rangle$ is a group.    I have to apologize if you find the argument needlessly long; I've been having some trouble with assuming things I am not supposed to assume (like what I'm going to prove), so I wanted to stay clear of that.
7599
This problem is from Herstein's 'Topics in Algebra'. I've thought about it a bit but haven't come up with much.  Let $G$ be a non-empty set with an associative product which also satisfies:   $\exists e\in G $ such that $\forall a \in G$, $a \cdot e=a$. Given $a \in G$, $\exists y(a) \in G$, such that $y(a) \cdot a =e$.   Prove that $G$ need not be a group.  I know that $G$ is a group if any one of those multiplications is switched around, i.e, either $a \cdot y(a)=e$ or $e \cdot a=a$. But I can't quite understand why in this case $G$ is not a group. I'd be much obliged if someone can give me a good explanation of why it must be and I'd also appreciate a counter-case if nothing else.   Thanks a lot. Cheers!
42168
Let $G$ be a non-empty finite set with an associative binary operation so that cancellation law holds, i.e. $ab=ac$ or $ba=ca$ implies $b=c$, for any choices of $a,b,c$ in $G$. Assume that there is an identity element $e$ in $G$. Show that $G$ is a group.  Proof: To show $G$ is a group, conditions must hold. Suppose that $ab=ac$, then $b=eb=(a^{-1}a)b =a^{-1}(ab)=a^{-1}(ac) =(a^{-1}a)c=ec=c$. So left cancellation holds in $G$.  2) Suppose $ba=ca$, then $b=be=b(aa^{-1}) =(ba)a^{-1}=(ca)a^{-1} =c(aa^{-1})=ce=c$. So right cancellation holds in $G$.  3) If $a$ exists in $G$ then $aa^{-1}=a^{-1}a=e$, where $a$ is an inverse of $a^{-1}$. Since inverses are unique, $(a^{-1})^{-1}=a$.  4) Let $x$ be the inverse of $ab$. Then $(ab)x=e$. By associativity we have $a(bx)=aa^{-1}$. Through left cancellation we have  $bx=a^{-1}bx=ea^{-1}=b(b^{-1}a^{-1})$ and $x=b^{-1}a^{-1}$.  Thus $(ab)^{-1}=b^{-1}a^{-1}$. So all conditions hold, $G$ is a group.  Is this proof correct? I know to show $G$ is a group these conditions have to be met. This is all I have to show right?
49135
The elements of $A_n$ is either of the form $(a,b,c,...)...$ or of the form $(a,b)(c,d)...$  In both cases, the element is a product of an even number of transpositions, not pairwise disjoint in the first case. I thought I would just show that any product of two transpositions is a product of two 3-cycles.  Consider the first case and let $\sigma = (v_1,v_2)(v_2,v_4)$ be one of the several products of two transpositions. Since $\sigma = (v_1,v_2,v_4) = (v_1,v_4,v_2)(v_1,v_4,v_2)$ showing that the product of two $\textit{not}$ disjoint transpositions is a product of two 3-cycles. Since the elements from the first case are all odd cycles, they can be expressed as a product of an even number of transpositions $\textit{not}$ pairwise disjoint. Thus the argument presented above could be done with all pairs of transpositions for the entire cycle.  Consider the second case and let $\sigma = (v_1,v_2)(v_3,v_4)$ again be one of the several products of two transpositions, this time disjoint. Clearly $\sigma = (v_1,v_3,v_2)(v_1,v_3,v_4)$, so I have shown that the product of two disjoint transpositions is a product of two 3-cycles. The elements from the second case are products of an even number of disjoint transpositions, so this second argument could be done for all those pairs.  This should cover all possible cases of elements in $A_n$ and thus it's generated by 3-cycles.  EDIT#1: Changed the argument to account for products of transpositions which are not disjoint.
24581
Suppose $F=\{0,a_1,\dots,a_{q-1}\}$ is a finite field with $q=p^n$ elements. I'm curious, why is the product of all elements of $F^\ast$ equal to $-1$? I know that $F^\ast$ is cyclic, say generated by $a$. Then the product in question can be written as  $$ a_1\cdots a_{q-1}=1\cdot a\cdot a^2\cdots a^{q-2}=a^{(q-2)(q-1)/2}. $$ However, I'm having trouble jumping from that product to $-1$. What is the key observation to make here? Thanks!
114711
Let $G = \{e, a_1, a_2, \dots, a_n\}$ be a finite abelian group and let $S$ be the set of all the elements of $G$ which are not equal to their own inverse. The set $S$ can be divided up into pairs so that each element is paired off with its own inverse. Prove that $(a_1 a_2 \dots a_n)^2 = e$.   Every $x$ in $G$, outside $S$ abides by $x^2 = e.$   Let $a_i$ be in $S$. Since $G$ is abelian, we have  $a_ia_i^{-1} = a_i^{-1}a_i$  $e = a_i^{-1}a_i$  Since $a_i^{-1} = a_i$, $a_i^2 = e.$  Thus, $\dots a_i^2a_j^2\ldots a_n^2 = e.$  Does that make sense?  edit:  We need to prove $a_1a_2...a_na_1a_2...a_n = e$, right? Let $a_i$ be in $G \setminus S$. Then $(a_i)^2$ is $e$, is it? If we rearrange the elements on the left side of the equation above according to what belongs to $S$ and what belongs to $G \setminus S$ and take the product of all the $(a_i)^2$, then every element belonging to $G \setminus S$ will  be cancelled in the right side of $a_1a_2...a_na_1a_2...a_n = e$. So, we only need to consider the elements in $S$. Suppose $a_j \in S$. Let $a_k$ be the inverse of $a_j$. We need to prove $(a_j)^2 \cdot (a_k)^2 = a_j \cdot a_j \cdot a_k \cdot a_k = e$. Since $S$ is abelian , $a_j \cdot a_j \cdot a_k \cdot a_k$ can be rearranged as $a_j \cdot a_k \cdot a_j \cdot a_k$ which is $e$.
90563
I am trying to prove the following and am looking toward the math.stackexchange community to comment on whether I am on the right track or not. Thank you in advance.    Let $n \geq 1$ be an integer. Suppose $\{G_1,...,G_n\}$ is a finite set of groups. Define the set $G = G_1 \times G_2 \times ... \times G_n$ of ordered n-tuples $(g_1,g_2,...,g_n)$ where $g_i \in G_i$. Show that the operation $\bullet$: $G \times G \rightarrow G$ given by $(g_1,g_2,...,g_n) \bullet (g'_1,g'_2,...,g'_n) = (g_1g'_1,g_2g'_2,...,g_ng'_n)$ makes $G$ into a group.    Proof of associativity: $((g_1,g_2,...,g_n) \bullet (g'_1,g'_2,...g'_n)) \bullet (g''_1,g''_2,...,g''_n)$ = $(g_1g'_1,g_2g'_2,...,g_ng'_n) \bullet (g''_1,g''_2,...,g''_n) $ = $(g_1g'_1g''_1,g_2g'_2g''_2,...,g_ng'_ng''_n)$ = $(g_1,g_2,...,g_n) \bullet (g'_1g''_1,g'_2g''_2,...,g'_ng''_n) = (g_1,g_2,...,g_n) \bullet ((g'_1,g'_2,...,g'_n) \bullet (g''_1,g''_2,...,g''_n))$  \  Proof of the existence of an identity For each $G_i \in G$, take its identity element $e_i$ and multiply it as follows: $(g_1,g_2,...,g_n) \bullet (e_1,e_2,...,e_3)$ = $(g_1,g_2,...,g_n)$.  \  Proof of the existence of an inverse: Since each $G_i \in G$ is a group, simply takes its multiplicative inverse $g_i^{-1}$ and multiply it: $(g_1,g_2,...,g_n) \bullet (g_1^{-1},g_2^{-1},...,g_n^{-1}) = (e_1,e_2,...,e_n)$.