I. 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) Give $a\in G$, there exists an element $a^{-1}\in G$ such that $a\cdot a^{-1}=e$.  Prove that $G$ must be a group under this product.  II. Prove, by an example, that right indentity element and left inverse does not imply that $G$ is group.  My solution:      I. Since $G$ is closed  set under an associative product, i.e. if $a,b,c\in G$ then $(a\cdot b)\cdot c=a\cdot (b\cdot c)\in G$. Taking $c=e$ we get $(a\cdot b)\cdot  e=a\cdot (b\cdot e)=a\cdot b \in G$. We have shown that $\cdot$ is binary operation.       Since $a\in G$ then $a^{-1}\in G$ and we have the following identities $$a^{-1}=a^{-1}\cdot e=a^{-1}\cdot (a\cdot a^{-1})=(a^{-1}\cdot a)\cdot a^{-1} $$   Then    $$ \begin{align} e &amp;= a^{-1}\cdot (a^{-1})^{-1} \\ &amp;= \left( \left( a^{-1}\cdot a \right) \cdot a^{-1} \right) \cdot \left( a^{-1}\right)^{-1} \\ &amp;= \left( a^{-1}\cdot a \right) \cdot \left( a^{-1}\cdot \left( a^{-1} \right)^{-1} \right) \\ &amp;= \left( a^{-1}\cdot a \right) \cdot e \\ &amp;=a^{-1}\cdot a.  \end{align} $$   Thus we have shown that    $$ a\cdot a^{-1}=a^{-1}\cdot a=e. $$   Then we see that    $$ e\cdot a= \left( a\cdot a^{-1} \right) \cdot a = a \cdot \left( a^{-1} \cdot  a \right) = a \cdot e = a. $$       We have shown that for this set $G$ and the associative binary operation assumed to be defined on $G$, the properties of the existence of a two-sided identity element in $G$ and the existence  in $G$ of a two-sided inverse for each element of $G$ are satisfied. Therefore $G$ is indeed a group.   II. But II indeed is true. Lets take the set $G=\{a,b,e\}$ and define the product $\cdot$ by the following identities: $e\cdot e=a\cdot e=b\cdot e=e$ and $a^{-1}=b, \ b^{-1}=a$ and consider the following multiplication table for our set $G$  $\begin{array}{c | c c c c c} \hline\hline  &amp; e &amp; a &amp; b  \\ \hline e &amp; e &amp; b &amp; b &amp; \\ a &amp; a &amp; a &amp; e &amp; \\ b &amp; b &amp; e &amp; a &amp; \\ \hline  \end{array} $  It's easy to verify that conditions of second problem hold for our $G$, however, $G$ is not group since we can show that $b=a$.  Is my reasoning above correct?  EDIT: Maybe this is a duplicate but I would like to know if my solution is true since I have solved it by myself. Especiaaly I am interested in the solution of the second problem.
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?
79458
Following the advice given in , I have started to study Group Theory from the very basics. My reference text is Abstract Algebra by Dummit and Foote. While going through the exercises (page 24) I found one problem which required more effort than others:  Assume $G = \{1, a, b, c\}$ is a group of order $4$ with identity $1$. Also assume that $G$ has no elements of order $4$. Use the cancellation laws to show that there is a unique group table for $G$. Deduce that $G$ is abelian.  I proceeded with solution as follows. Since $G$ is of even order there must be at least one element of order $2$ (this statement itself was one of the problems in the exercises and is easy to handle). From the wording of the question it seems that none of the elements $a, b, c$ is singled out with specific properties and hence $a, b, c$ must behave in exactly the same manner. Hence each of them is of order $2$. Thus $a^{2} = b^{2} = c^{2} = 1$.  Next I analyze product $ab$. Clearly it can't be $a$ (as $b$ is non-identity), or $b$ (as $a$ is non-identity) or $1$ (because $ab = 1 = aa$ implies $b = a$). Hence $ab = c$ and similarly $ba = c$. By the similar nature of all elements $a, b, c$ it follows that $bc = cb = a$ and $ca = ac = b$. Thus the operation of the group is defined properly for all elements and clearly it is abelian.  I think my solution is OK, but I am not sure. Note that I have not used the fact that $G$ has no elements of order $4$. Please let me know :  1) if my solution is correct. If not then point out the flaws.  2) if there is any better / shorter solution. If so provide hints and not solution.  The question may sound too easy / simple but I request to treat me like a beginner who has never heard of Group Theory and is reading fresh from the reference text I mentioned. Thanks in advance for your inputs.  EDIT: Thinking further about this problem I wondered what would happen if an element of order $4$ was allowed in $G$. In that case I think that the group has to be isomorphic to the group $G_{1} = \{1, 2, 3, 4\}$ with modulo $5$ multiplication as the group operation. The argument is as follows. Suppose $a$ is of order $4$ and let $b$ be its inverse then $b$ is also of order $4$. Now $c$ has to be its own inverse so that $c^{2} = 1$. Again $a^{2} \neq 1$ (as $ab = 1$), $a^{2} \neq a$ (as $a \neq 1$), $a^{2} \neq b$ (as it would mean $b^{2} = 1$ and thus $b$ would be of order $2$). Hence $a^{2} = c$ and similarly $b^{2} = c$. Next we can see that $ac \neq a$, $ac \neq c$, $ac \neq 1$ (as $ab = 1$) so that $ac = b$. Similarly $ca = b$, $bc = a$, $cb = a$. Also it is clear that this group turns out to be cyclic with both $a$ as well as $b$ as generators. And it also follows that there are the only two ($G$ in original question and $G$ in new variation) groups of order $4$ upto isomorphism.  I hope I have learnt something from the answers given for the original question and this solution is correct. Please let me know if there is any problem with this reasoning.
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?
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.
75204
Problem:  Let T be a nonempty set and A(T) the set of all permuations of T. Show that A(T) is a group under the operation of composition of functions.  Permutation of the set T is the bijective function $f: T \rightarrow T $  For A(T) to be a group it must be closed, associate have an identity for every element and an inverse.  Since f is  bijection the composition of two bijective functions is itself bijective so A(T) is closed. Every bijective function has an inverse, and composition of functions is associative. What about the identity? Are we to take $ A(T) = S_n $ here, which would be assuming that $T$={$1, ..., n$}? In which case $ I = \begin{bmatrix}1&amp;2&amp;...&amp;n\\1&amp;2&amp;...&amp;n\end{bmatrix} $  Not sure what I'm supposed to be showing since $T$ and A(T) are arbitrarly specified here.
8302
My reference for learning abstract algebra says that in an abelian group $G$ the product of two elements (say $a$ with order $n$ and $b$ with order $m$) has order $mn$ if $gcd(n,m) = 1$.   What I don't understand is where the $gcd(m,n)=1$ comes into use in proving this fact? For example, $ a^{mn} = (a^n)^m = (e)^m = e$ (identity element). A similar argument for $b^{mn}$ together with the fact that $G$ is abelian is enough to show that $(ab)^{mn} = e$. From here, how do I conclude for an integer $r &lt; mn$ that $(ab)^r \neq e$  Also, are there any other conclusion that can be drawn if $gcd(n,m) \neq 1$ ?  Thanks for any feedback in advance !
29831
Closure under multiplication:  $\to$  Let $a, x, y \in H.$  Since $H$ is closed under inverses and multiplication, $xax^{-1} \in H$ and $yay^{-1} \in H.$ Since $H$ is closed under multiplication, $xyax^{-1}y^{-1} \in H.$   $\leftarrow$  Suppose $xyax^{-1}y^{-1} \in H$, then $a \in H$ since $H$ is closed under multiplication. So, $xy \in K.$  Closure under inverses:  Since $H$ is closed under multiplication and inverses, if $a \in H,$ then $x^{-1}ax \in H.$ If $x^{-1}ax \in H,$ then $a \in H$ because $H$ is a group. So, $x^{-1} \in K.$  Identity:  Suppose $a \in H.$ Then $a = eae^{-1} \in H.$ So, $e \in H.$  I'd like to see how I can fix/improve this proof.
6360
What is an instructive example of a set $X$ equipped with two monoid structures $(X,+,0)$, $(X,\cdot,1)$, such that $+$ is commutative, the distributive laws hold, but $0 \cdot x = 0$ or $x \cdot 0 = 0$ do not hold?  Notice that in case these two absorbing laws hold, one calls $(X,+,0,\cdot,1)$ a semiring. At first sight, it might be surprising that these laws have to be imposed, but this is quite natural from a more general point of view, namely the multiplication $\cdot : X \times X \to X$ should be an additive monoid homomorphism in each variable, and it is known that one has to demand that monoid homomorphisms preserve the neutral element.
15672
Suppose $G=\{a_1,...,a_n\}$ is a finite abelian group, and let $x=a_1a_2\dotsm a_n$.   Prove that if there is more than one element of order $2$ then $x=e$.   What I've done so far: (#1 is just for illustration, There should be at least $3$ such elements)   $G$ has an even number of elements with order $&gt;2$ (They are paired so: $(a,a^{-1})$ ). And then there are $e$ and $\{b|o(b)=2\}$. Also, if $G$ has at least one element of order 2 then $|G|$ is even (Lagrange). Thus $|\{b|o(b)=2\}|$ must be odd. If there is exactly one element $b$ of order $2$ then $x=b$. That's because $G$ is abelian, and we can write: $x=e\cdot b\cdot(a_1 a_{1}^{-1}\dotsm a_n a_{n}^{-1})=b\cdot(e\dotsm e)=b$. If there are exactly $3$ such elements $a,b,c\in G$, then, as shown above, $x=abc$ and thus $x=e$ (That's because $(abc)^2=a^2b^2c^2=e$ which means $abc\in \{a,b,c,e\}$. If $abc=a$ then $b=c^{-1}=c$ which is false, same goes for $abc=b,abc=c$.)   I haven't managed to find a reason why the claim must hold for $5, 7, ...$ (i.e. All other odd integers). Does anyone have an idea?  Thanks in advance. Please excuse my English.  edit: Not a canonical answer but a simple one (if such even exists).
100174
I'm just wondering the reason behind being able to do this:  $$\gcd(a,c) \cdot \gcd(b,c) = \gcd(ab,bc,ac,cc)$$  Is there an elementary proof to the reasoning behind this?
116030
I know how to prove this using an element and subset proof but that's not what I'm being asked to do. I'm suppose to prove it using the distributive law and the associative law but I'm just not seeing how to do this. Can you give me some advice on how to start this. I figure the first thing would be $$((A \cup B) \cap A) \cup ((A \cup B) \cap B^c))$$ But I'm not sure where to go from there.
106485
Let $G$ be a finite group and suppose there exists $f\in\text{Aut}(G)$ such that $f^2=\text{id}_G$, i.e., $f$ is its own inverse, and such that $f$ has no fixed points other than the identity $e$ of $G$, i.e., $f(x)=x\Rightarrow x=e$. Show that $G$ is necessarily abelian.   While trying to do this exercise I noticed two facts.  First, $g$ and $f(g)$ have the same order because $o(f(g))|o(g)$ and, applying $f$ again and using $f^2=\text{id}_G$, $o(f(f(g)))=o(g)|o(f(g))$ and, once the order of any element is $\geq 1$, it follows that $o(g)=o(f(g))$. Also, it's easy to see that $g$ and $f(g)$ commute.  Second, there cannot exist such an automorphism if the order of $G$ is even, because $f(e)=e$ and we can form pairs like $\{g,f(g)\}$ with $f(g)\neq g$ that are invariant under $f$, i.e., $f(\{g,f(g)\})=\{g,f(g)\}$. But once the order of $G$ is even, proceeding with the construction of the pairs, we'll end up with just one element $\neq e$, so we must have some $\gamma\in G\setminus{e}$ with $f(\gamma)=\gamma$, contradicting the hypothesis.  Let $n$ be the order of $G$ and let's fix an enumeration of the elements of $G$, say $G=\{g_1,\ldots,g_n\}$. My approach was the following. For each $\sigma\in S_n$, let $x_{\sigma}=\prod_{i=1}^ng_{\sigma(i)}$. Then we have that $f(x_{\sigma})=x_{\sigma^{\prime}}$. If it's shown that $f(x_{\sigma})=e\;\forall\;\sigma\in S_n$, then we'd have $x_{\sigma}=e\;\forall\;\sigma\in S_n$, which implies that $G$ is abelian. The problem is that $S_n$ has $n!$ elements while $G$ has $n$ elements, so there repetitions among the $x_{\sigma}$, and we cannot apply directly the reasoning of the last paragraph.  Is this the right way, or is there an easier manner to solve this?
107609
Let $(S, \cdot)$ be a semigroup. There are natural numbers $p,q \geq 2$ such that $a^pb^q=ba$ for all $a,b \in S$. Prove that $S$ is commutative.   I wrote   $$\begin{align} a^{p+1}b^{q+1} &amp;=b^{(q+1)p}a^{(p+1)q} \\  &amp;=b^{p}\cdot(b^q)^p \cdot (a^p)^q\cdot a^q \\  &amp;=b^p\cdot a^p \cdot b^q \cdot a^q \\  &amp;= b^p\cdot b \cdot a \cdot a^q \\  &amp;=b^{p+1}a^{q+1}. \end{align}$$  From the given identity I also got $a^{p+1}b^{q+1}=abab$. Using $a^{p+1}b^{q+1}=b^{p+1}a^{q+1}$ I then got $abab=baba$.  Making $a=b$ in the statement gives $a^{p+q}=a^2$. I don't know what to do from there.
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: $(g_1,g_2,...,g_n) \bullet (g_1^{-1},g_2^{-1},...,g_n^{-1}) = (e_1,e_2,...,e_n)$.
16266
If $G$ is a group show that if $(a \cdot b)^2 = a^2 \cdot b^2$ then $G$ must be abelian.   $\begin{aligned}(a \cdot b)^2 = a^2 \cdot b^2 &amp; \iff (a\cdot b)\cdot(a \cdot b) = (a \cdot a)\cdot (b \cdot b) \\&amp; \iff a \cdot (b \cdot (a \cdot b)) =a(a \cdot (b\cdot b)) \\&amp; \iff (a^{-1} \cdot a) \cdot (b \cdot (a \cdot b)) =(a^{-1}\cdot a)(a \cdot (b\cdot b))\\&amp; \iff (b \cdot (a \cdot b)) =(a \cdot (b\cdot b)) \\&amp; \iff (b \cdot a) \cdot b =(a \cdot b)\cdot b\\&amp; \iff (b \cdot a) \cdot (b \cdot b^{-1}) =(a \cdot b)\cdot (b \cdot b^{-1}) \\&amp; \iff b \cdot a = a \cdot b\end{aligned}$  Thus $G$ must be abelian. Is this right? Is there less clunky way to write it if it's?
7251
Suppose we have a set  $S=\{a,b\}.$ Obviously, the total number of binary operations on $S$ is the number of all the pairs such that $$*(a,b)=a \, \text{or}  \, b,$$ which gives  $ 2 \cdot 2 \cdot 2 \cdot 2=2^4$ possible ones. Is there a way to find which of these $16$ binary operations are associative without writing all the operation matrices and checking it by hand?
44145
An exercise from Chapter 20 of "How to Think Like a Mathematician" by Kevin Houston:     Let $n$ be an odd positive integer. Let $(a_1,a_2,\dots,a_n)$ be an arbitrary arrangement (i.e., permutation) of $(1,2,\dots,n)$. Prove that the product $(a_1-1)(a_2-2)\cdots(a_n-n)$ is even.   For example, for $n = 3$, we can have $(a_1, a_2, \dots, a_n) = (3, 1, 2)$, so this yields that $(3-1)(2-1)(1-2)$ is even.  Would the following be considered a full solution?  Each even an has to be paired with a ($a_n$ -odd) in order for each even an to become odd. However after pairing them we have one -odd left as there is one more odd than even in the set $(1,2,\dots,n)$ where $n$ is odd. This must be paired with a odd number in a bracket. As odd-odd is even (the proof of this is trivial), then one of the bracket must be even so the product is even.  Would THIS solution be considered full?