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!
52872
Hungerford's Algebra poses the question: Is it true that a semigroup $G$ that has a left identity element and in which every element has a right inverse is a group?  Now, If both the identity and the inverse are of the same side, this is simple. For, instead of the above, say every element has a left inverse. For $a \in G$ denote this left inverse by $a^{-1}$. Then  $$(aa^{-1})(aa^{-1}) = a(a^{-1}a)a^{-1} = aa^{-1}$$  and we can use the fact that   $$cc = c \Longrightarrow c = 1$$   to get that inverses are in fact two-sided:  $$ aa^{-1} = 1$$  From which it follows that  $$a = 1 \cdot a = (aa^{-1})a = a (a^{-1}a) = a \cdot 1$$   as desired.  But in the scenario given we cannot use $cc = c \Longrightarrow c = 1$, and I can see no other way to prove this. At the same time, I cannot find a counter-example. Is there a simple resolution to this question?
50076
I had this "almost bonus" question on the final in Group Theory recently: prove that there is no such group $G$ which would satisfy Aut$(G)\cong \mathbb{Z}_n$, where $n$ is an odd integer. I don't have much certainty if this proof is OK, and one's opinion would be appreciated.  Here's my attempt:     Suppose that there exists such a group $G$ that satisfies the above condition. Since $\mathbb{Z}_n$ is cyclic, Aut$(G)$ is also cyclic and $G$ is abelian, which implies that Inn$(G)\cong \{e\}$. Thus Aut$(G)=$Out$(G)$. (I don't think this fact is important here though). Now, since an automorphism sends a generator to a generator, and since each automorphism is completely determined by such mapping, |Aut($G$)| must be of factorial order. But an integer factorial is always even. However, $|\mathbb{Z}_n|$ is odd, which is a contradiction. Therefore, no such $G$ exists.   I'm afraid that my proof is not very nice, but at least a genuine attempt was made.
61758
I am trying to demonstrate that  $$((a,b),c)=(a,(b,c))\qquad\text{ (Associativity of GCD)}.$$   My proof is the following:  Since $((a,b),c)|(a,b)$, $(a,b)|b$, $(a,b)|a$ and by transitivity we have $$((a,b),c)|b, ((a,b),c)|a.\tag{1}$$ We also have $((a,b),c)|c$.   By linearity we have $\forall x,y\in \mathbb{N}$ $$((a,b),c)|bx+cy.\tag{2}$$  We can choose $x$ and $y$ so that $$bx+cy=(b,c).\tag{3}$$   Therefore $((a,b),c)|(b,c)$ and $((a,b),c)|a$.   Again by linearity $$((a,b),c)|xa+y(b,c)\tag{4}$$ and we can choose $x$ and $y$ so that $$xa+y(b,c)=(a,(b,c)).\tag{5}$$   This mean $((a,b),c)|(a,(b,c))$.   Using an analogous argument we can show that $$(a,(b,c))|((a,b),c)$$ and by the reflexivity the thesis follows.      Is this correct?
30370
Let $G$ be a group and let $\mathrm{Aut}(G)$ be the set of all automorphisms of $G$. Prove that $\mathrm{Aut}(G)$ is a group under the operation of composition for functions.    I am not too sure if I am proving $\mathrm{Aut}(G)$ itself is a group. Would this mean then that I would have to make sure it fulfills all the group requirements? If so I am not too sure how to go about it. And to prove closure, I would think I need to pick say $f,g \in \mathrm{Aut}(G)$ then use the fact that these are automorphisms. From here, I am not too sure how to proceed.
2008
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.
6998
I am interested in the following question:  Does $G \times H \cong G$ necessarily imply that $H$ is trivial?  In order to demonstrate it is not, I attempted to construct an example, but I am not entirely sure it is valid:  Let $G = \langle \mathbb{Z}; [a,b] \ (\forall a,b \in \mathbb{Z})\rangle$.  Then $G \times G = \langle \mathbb{Z} \cup \mathbb{Z}; \ \{[a_1, b_1]\} \cup \{ [a_2, b_2]\} \cup \{[a_1,b_2]\}\rangle$, where my notation is intent to convey that the direct product presentation has the abelian relations from the "first" copy of G, the abelian relations from the "second" copy of G, and the required abelian relations which commute generators of the "first" copy with those of the "second" copy.    In this example, since our cardinalities are equal, $|\mathbb{Z} \times \mathbb{Z}| = |\mathbb{Z}|$, and since this implies that commuting all elements of $G$ and commuting all elements of the same cardinality generating set $G \times G$ requires a bijection between the relator sets, then we should have a bijective group homomorphism $\phi : G \rightarrow G \times G$ so that we have isomorphism, but obviously $G$ is not trivial.  Therefore this provides the necessary counterexample to the original statement.  My first question is: is there anything wrong with this proof?  My second question is: if there is nothing wrong with the proof, is there a more standard way of constructing this type proof?  In general, it would seem that whenever you have isomorphic generating sets and isomorphic relator sets, then this should be enough to describe any number of counterexamples to the claim that $H$ must be trivial.  Is this true?  Finally, since direct products can be viewed as the products in the category Ab, and since all final objects must be isomorphic, would there be an easier way (or maybe just an insightful way) of construct this argument entirely in a category theoretic context?  Thanks!
62301
Let $G$ be a set with a binary operation *, associating to each pair of elements $x$ and $y$ of $G$ a third element $x*y$ of $G$. Suppose that the following properties are satisfied:   $(x*y)*z = x*(y*z)$ for all elements $x$, $y$, and $z$ of $G$ (the Associative Law); there exists an element $e$ of $G$ such that $e*x = x$ for all elements $x$ of $G$; for each element $x$ of $G$ there exists an element $x'$ of $G$ satisfying $x'*x = e$.   How to prove  i. $x*e=x$ and  ii. $x*x'=e$,  so that $G$ is a group?
24118
If the order of all nontrivial elements in a group is 2, then the group is Abelian. I know of a proof that is just from calculations (see below). I'm wondering if there is any theory or motivation behind this fact. Perhaps to do with commutators?    Proof: $a \cdot b = (b \cdot b)  \cdot (a \cdot b) \cdot (a \cdot a) = b \cdot (b \cdot a) \cdot (b\cdot a) \cdot a = b \cdot a$.
36424
While reading Cours d'algèbre by D. Perrin, I found the following claim:     Proposition. If $G$ is a group, then $D(G)\subseteq G^2$, where $G^2$ is the subgroup generated by the squares of $G$.   I understand that it suffices to prove that for all $x,y\in G$, $[x,y]:=xyx^{-1}y^{-1}$ is a product of squares.  Equivalently, since $G^2$ is characteristic, hence normal, it suffices to establish that $G/G^2$ is abelian.  For now, none of my attempts have been fruitful. Any enlightenment will be greatly appreciated!
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.
35489
I'v found this proposition.  If $G$ is a finite  group , $ A \subset G $ a subset and $|A| &gt; \frac{|G|}{2} $ then $AA = G $.  Why this is true ?
59108
Let $𝐺$ be the group generated by the matrices $𝐴=\begin{bmatrix}2&amp;0\\0&amp;1\end{bmatrix}$, $𝐵=\begin{bmatrix}1&amp;1\\0&amp;1\end{bmatrix}$, where the group operation is matrix multiplication.      Prove that $𝐺$ contains a subgroup $𝐻_𝑖$ for each $𝑖\in \mathbb{Z}$ such that   $$ \cdots \subset H_{-2}\subset H_{-1} \subset H_0 \subset H_1 \subset \cdots$$   So this is all that is given and i'm not sure where to start from this question. I know that $ABA^{-1}$ = $B^{2}$. From this we can replace A with $A^2$ and so on but i am not sure how i would prove this. Can someone help please?
84773
I was trying to find an example of a group $G$ and normal subgroups $N_1 , \ldots , N_n$ such that $G = N_1\cdots N_n$ and $N_i \cap N_j = \{e\}$ for all $i \neq j$ and yet $G$ is not the internal direct product of  $N_1, \ldots , N_n$ .  It is easy to show that G is the internal direct product of normal subgroups $N_1 , \ldots , N_n$ iff  (i) $G = N_1\cdots N_n$  (ii) $N_i\cap (N_1 \cdots N_{i-1}\cdots N_{i+1} \cdots N_n) = \{e\}$ for $i = 1, 2, \ldots, n$.  So I guess the question boils down to produce an example of a group $G$ and its normal subgroups  $N_1 , \ldots , N_n$ such that   (1) $G = N_1\cdots N_n$  (2) $N_i \cap N_j = \{e\}$ for all $i \neq j$   but     (3) $N_i \cap (N_1 \cdots N_{i-1} N_{i+1} \cdots N_n) \neq \{e\}$ for at least one $1 \le i \le n.$ Thanks in advance for help....
55572
Motivation: If operation is change, and there is no change, then there is no operation. Then the meaning of  1+0 =1 is not in the equality itself but in the comparison with other equalities as  1+2=3. The following definition does not use the meaningless equality  1+0 =1, or generally a*e=a. I guess that trying to prove statements using the next definition, matters/problems will arise that do not arise with the standard definition. Let be the non empty set G = {a,b,c,d,.....z} with the operation *, satisfy the following 3 conditions:  a(bc) = (ab)c, for every a, b, c ∈ G (associative law). For every a,b there is c (unique) such that ac = b. (c  is the &quot;path&quot; from a to b). For every a,b there is d (unique) such that bd = a. (d is the &quot;path&quot; back from b to a).  P.S.: the meaning of 2) and 3) is the possibility to &quot;go&quot;/&quot;move&quot; from any member to any member, through the same path (if a-&gt;b-&gt;c then c-&gt;b-&gt;a as in the Klein group) or through other path (if a-&gt;b-&gt;c then c-&gt;X-&gt;a). Prove: G,*  is a group. Proof: Associativity: From 1) Identity element: Let's prove for every a there is e such that ae=ea=a. Proof: According to 2) For every a,a there is e such that ae = a. (3) According to 2) for every e,a there is x such that ex=a, that means ae =ex =a (4) Let's proof x=a. Multiply by a on the left, such that the result is some g ∈ G: ae = ex = a (4) &lt;---&gt; (a)ae = (a)ex = aa =g &lt;---&gt; aae = (ae)x = aa =g (associative) &lt;---&gt; &lt;---&gt; aae = ax = aa =g (3). (5) According to 2) for every a,g there is x (unique) such that aa=g and ax=g , then a and x are the same, and it's proved x=a. Then replacing in (5): ae = ea = a. Inverse element: Let's prove that for every a, there is b such that a ⋅ b = e and b ⋅ a = e According to 2) for every a,e there is b such that ab = e. (6) According to 2) for every b,e there is X such that bX=e. (7) Let's proof that X=a ab = bX = e (6),(7) Multiply be a on the right: aba =ea a(ba) = ea; Associativity ea = a; Identity element a(ba) = a; then ba = e bX=e according to (7) ba = bX = e; according to 2) for every b,e there is x unique such that bX = e then x=a and it's proved that for every a, there is b such that a ⋅ b = e and b ⋅ a = e.
3922
G is the set of matrices of the form $G=$$\begin{pmatrix} x &amp; x \\ x &amp; x \end{pmatrix}$. So for this set to be a group I know it needs to be:    Closed under matrix multiplication The Associative Property holds Contains an Identity Element Every element needs to have an inverse   So the form of the matrices is such that all the elements are the same but not 0. How do I go about proving these?  Working through this problem, I seem to have hit a contradiction. Since G is a subgroup of the bigger $2x2$ nonsingular matrices group why does G not have the same identity element as its parent group? Namely \begin{bmatrix} 1 &amp; 0 \\ 0 &amp; 1 \end{bmatrix}  Isn't the subgroup supposed to have the same identity element as its parent group?
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: $(g_1,g_2,...,g_n) \bullet (g_1^{-1},g_2^{-1},...,g_n^{-1}) = (e_1,e_2,...,e_n)$.
66551
A quick preface to the below question: this is my first post on math.se! I am excited to begin participating in such a wonderful community. Any feedback as to how I can improve this or subsequent posts is certainly feedback from which I could benefit.      (Exercise 3.2.2 of Tao's Analysis I) Use the  (and the singleton set axiom, which guarantees the existence of the singleton set) to show that if $A$ is a set, then $A\notin A{. \kern 0.05em}^\text{1}$ Furthermore, show that if $A$ and $B$ are sets, then $A\notin B$ or $B\notin A$.    Thoughts: Consider some set $A$ and the singleton $\{A\}$. By the axiom of regularity, the only element of $\{A\}$, namely $A$, is not a set or disjoint from $\{A\}$. Clearly, $\{A\}\cap A=\emptyset$ and thus, $A\notin A$.   Moreover, suppose $B$ is some other set such that $\neg(A\notin B \lor B\notin A)$, i.e. $A\in B\land B\in A$. This implies, in very vauge notation, that $A=\{\{A,\dots\},\dots\}$ and $B=\{\{B,\dots\},\dots\}$. Can I get a contradiction out of this? In the case of $A$, we get that $x\in\{A,\dots\}\lor x\in \{a\}$ for some $x\in A$ and all other objects $a\in A$. How can I prove that there exists no element of $A$ (and $B$) that is not a set unless it is disjoint from $A$. I would greatly appreciate any hints as to how I can complete my argument.     1 This first problem has been discussed on this site times before (see  and ). I include a solution only because the context demands it, but to avoid a duplicate question, I ask that it not be discussed here.