Let $(S,*)$ be a finite semigroup with identity. Prove that $S$ is a group iff $S$ has only one element $x$ such that $x^2=x$. Attempt: Does this approach true? $(\Rightarrow)$ Let $S$ be a group. Denote $e_S$ be an identity element in $S$. Clearly, $e_S$ is idempotent since $e_S^2 = e_S$. Suppose that there is another idempotent in $S$, say $x$. Then, $x^2 = x$. So, \begin{equation*} x^2=x \Leftrightarrow x^{-1}x^2 = x^{-1}x \Leftrightarrow x = e_S. \end{equation*} Hence, $x=e_S$. Thus, the only idempotent in $S$ is $e_S$. $(\Leftarrow)$ First, to show for any $x \in S$, there exist $n \in \Bbb N$ such that $x^n$ is idempotent. I found this proof in the ProofWiki. Next, since the only idempotent in $S$ is $e_S$, then $x^n$ must equal to $e_S$. I have proved this already. But, I am still confused. If $n=2$, then $a^2 = e_S$ i.e. the inverse of $a$ is itself. But, what about $n=3$? $a^3=e_S$, does it implies that the inverse of $a$ is $a^2$?
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Let $G$ be a finite semi-group with identity such that it has only one idempotent.Is $G$ a group?  It only remains to show that for any $a\in G$ $\exists b\in G$ such that $ab=ba=e$ where $e$ is the identity of $G$  Also $e$ is the only idempotent of $G$ .How to proceed next?
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Let $G$ be a group such that for all $x\in G$ we have $x^2=e$. Show that if $G$ is finite then the order of $G$ is $2^n$.  Here is the solution I have seen in a book.      If G is finite, it can be considered as a vector space over the field $\Bbb{Z}/2\Bbb{Z}$, and then necessarily is finite dimensional, which gives $G$ as a vector space isomorphic to $(\Bbb{Z}/2\Bbb{Z})^n$.   I don't understand why $G$ can be considered as a vector space over the field $\Bbb{Z}/2\Bbb{Z}$. Can, someone, explain this? Please.
108605
Let $G$ and $G'$ be groups. Let $\phi: G \to G'$ be a homomorphism. Let $H$ be a group of $G$ and let $H'=\{\phi(h): h\in H\}$.  Want to show that $H'$ is a subgroup.  Proof:  (i)  $H'$ has identity since $\phi(e_G)\in H'$.  This is because $\phi(e_G)=e_{G'}$.  (ii)  $H'$ is closed because of the following: Let $h_{1}^{'}, h_{2}^{'} \in H'$.  Then there exists an $h_1,h_2$ in $H$ such that $h_1^{'}=\phi(h_1)$ and $h_2=\phi(h_{2}^{'})$.  Then $h_1^{'}h_2^{'}=\phi(h_1)\phi(h_2)=\phi(h_1h_2)$.  (iii) For each $h^{'} \in H'$ there exists an $h \in H$ such that $\phi(h)=h'$.  But, $h'^{-1}=\phi(h)^{-1}=\phi(h^{-1})$.  Hence it has inverses.  Did I prove this correctly?
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I'm stuck at this question. Can someone please help me?     Prove that if a group contains exactly one element of order 2, then that element is in the center of the group.   Let $x$ be the element of $G$ which has order 2. Let $y$ be an arbitrary element of $G$. We have to prove that $x \cdot y = y \cdot x$.  Since $x$ has order $2$, \begin{equation} x^2 = e \end{equation} That is, \begin{equation} x^{-1}=x \end{equation}  I don't really know how to proceed. I've tried a number of things, but none of them seem to work.
13548
Let $G$ be a finite group having order $n$, prove that $n$ is odd if and only if for each $a\in G$, there is $b\in G$ such that $a=b^2$. I first assume $n$ is odd, then let $a$ be an element in $G$, since $a^n=e$, we have $$a=a^{n+1}=(a^{(n+1)/2})^2$$ which tells $a$ is the square of $a^{(n+1)/2}$.  To prove the converse, I use contraposition, which I assume $n$ is even, then I want to find $a\in G$ such that it is not a square. My intuition is, "assume" we can always find an element $a\in G$ such that the smallest positive integer $k$ satisfy $$a^k=e$$ is $k=n$. Then it is not true that $a$ can be expressed as a square, indeed if $a=b^2$ for some $b\in G$, then the smallest $k$ satisfy $b^k=e$ will be $k=2n$, but this contradicts the smallest $k$ cannot exceed $n$.  My problem is, is the assumption I made always true? If it is true then it should be in the very beginning of the textbook, but I have forgotton it. Any help?  Source: Apostol Analytic Number Theory.
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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.
53662
I am trying to prove that any conjugate of $a$ has the same order as $a$.  Let $G$ be a group and let $a \in G$.  An element $b \in G$ is called a conjugate of $a$ if $b=xax^{-1}$.  My professor gave us a hint.  Prove $a^k=e$ iff $b^k=e$ for all $k \in \mathbb{Z}$ implies $o(a) = o(b)$.  My first move is to $b^k = (xax^{-1})^k = (x^{-1})^ka^kx^k = (x^{-1})^kex^k = (x^{-1})^kx^k=x^{-k+k}=x^0$  I am not sure that I am doing this correctly?  I am trying to get $b^k=e$.  However when I raise $(xax^{-1})^k$, I am not sure if I what I have done is correct?  I am kind of confused here.
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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?
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Please note: I posted this question about an hour ago on my account (I just made it as I am new to the site). I revisited my post and saw that I had not given definitions of what I was asking (I thought they were common notations) and it would not let me edit the post for some reason so I will ask here now and hopefully someone can delete the other thread.  How to prove for a finite group $G$ that $|\langle g \rangle|=o(g)$?  Here are the definitions:  Let $G$ be a finite group and $g \in G$  $o(g)$ is the smallest $m \in \Bbb{N}$ such that $g^m=e$ (the identity element of the group). (If such an $m$ exists)  $\langle g \rangle=\{g^m : m \in \mathbb{Z}\}$  I really don't know how to show that this is true I tried to say let $o(g)=m$ and then show that $|\langle g \rangle|$ has exactly $m$ elements but I wasn't even sure it did because it would contain the elements $g^0,...g^{m-1}$ so that would mean it's true but what about $g^{-1}$ and the other elements with negative coefficients so it seems like it would actually be bigger than $m$?  Thanks.
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Let $G$ be a finite group and $f:G\to G$ an isomorphism. If $f$ has no fixed points (i.e., $f(x)=x$ implies $x=e$) and if $f\circ f$ is the identity, then $G$ is abelian. (Hint: Prove that every element in $G$ has the form $x^{-1}\cdot f(x)$.)   With the hint, I can see that for all $t\in G$, $f(t)=t^{-1}$, and since $f$ is an isomorphism, $G$ is abelian. But I can't see how to prove the hint or why it's evident.  Source : Rotman J.J. Introduction to the theory of groups, exercise 1.50
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On  I asked examples of finite free objects. I got answer "The free idempotent semigroup satisfies $x^2=x$ for each element $x$. And I confirm it is finite if it is finitely generated." with reference to a paper.  If I understood correctly the concept of free object, then elements of free idempotent semigroup of three generators can be seen as strings like "abaca" with "non-repeating" property: do not count for example "abcbc" because has "bc" repeats. Well, on  it is said that there is infinitely many strings of three letters with that property.  So either at least one of the answers is wrong, or then I have understood this wrong.  If given semigroup is finite, how many elements it has?
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The proof goes on like this :   $Z$ denotes integers.  $a^m=e$ for some positive integer $m$. Let $n$ be the smallest positive integer such that $a^n=e$ if $s \epsilon Z$ and $s=nq+r$ for $0\le r &lt; m$ then $$a^s=a^{nq+r}=a^r \ \ \ (*)$$.  Then after this statement, the book proves that all elements of the finite cyclic group is distinct and creates a homorphism between this group and  $&lt;Z_{n},+_{n}&gt;$  What I couldn't get in the proof is what does $(*)$ tell us? I couldn't understand it and why is it needed in the proof? Also why do we work with positive integers, why do we let $n$ be the smallest integer instead of minimizing $|n|?$
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I tried to construct a Cayley table of an algebraic structure called inverse semigroup. No success so far. I just end up with more complicated structure (monoid, group). Thank you kindly. I may think, that this phrasing is more accurate and more to the point. Also, I have checked many sources and didn't find an example... One possibility would be, that it is not possible (which I doubt). Or why would it be not possible? One chance is to try for any set $X$, let $I(X)$ be the set of all partial bijections on $X \to X$, i.e. bijections between subsets of $X \to X$. The composite of partial bijections is their composite as relations (or as partial functions). In fact, any inverse semigroup is isomorphic to a sub-inverse-semigroup.
81509
Let $G$ be a group s.t. every non-identity element has order 2. If $G$ is finite, prove $|G| = 2^n$ and $G \simeq C_2 \times C_2 \times\cdots\times C_2$  I know G is abelian since $ab = (ab)^{-1} = b^{-1} a^{-1} = ba$ for all non-trivial $a,b \in G$ so I have several questions remaining:    How do I prove $|G| = 2^n$? I'd like to say we use induction to prove this but I'm at a loss as to where I would start. Why is $G \simeq C_2 \times C_2 \times\cdots\times C_2$?   I've read several answers to question similar to this yet unfortunately most of them involve Galois Fields and vector spaces, both concepts I'm unfamiliar with. I'd greatly appreciate an intuitive proof.
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Let $(G,\cdot)$ be a non-empty finite . Is there any $a\in G$ such that: $$a^2=a$$ It seems to be true in view of theorem 2.2.1 page 97 of  book (I'm not sure).  But is there an elementary proof?  Theorem 2.2.1. [R. Ellis] Let $S$ be a compact right topological semigroup. Then there exists an idempotent in it.  This theorem is also known as .
109664
This is Exercise 1.9.17 of Howie's "Fundamentals of Semigroup Theory".  The Details:  Let $S$ be a commutative semigroup.     Definition 1: Let $\theta_n^S$ for $n\ge 1$ be given by   $$a\theta_n^Sb \iff (\forall x\in S^n)\, xa=xb.$$      Lemma 1: The relation $\theta_n^S$ is a congruence on $S$ and $$\theta_1^S\subseteq\theta_2^S\subseteq \dots$$   Proof: For all $a\in S$, $xa=xa\forall x\in S^n$. Hence $a\theta_n^Sa$, so $\theta_n^S$ is reflexive.  Suppose $a\theta_n^S b$. Then, for $x\in S^n$, $xa=xb$, i.e., $xb=xa$, so $b\theta_n^Sa$. Hence $\theta_n^S$ is symmetric.  Suppose $a\theta_n^S b$ and $b\theta_n^S c$. Fix $x\in S^n$. Then $xa=xb=xc$, so $a\theta_n^S c$. Hence $\theta_n^S $ is transitive.  Suppose $a\theta_n^S b$ and $c\theta_n^S d$. Then for $x=x_1\dots x_n\in S^n$, $$\begin{align} x(ac)&amp;=(xb)c \\ &amp;=(x_1\dots x_{n-1})(x_nb)c \\ &amp;=(x_1\dots x_{n-1})(x_nb)d \\ &amp;=x(bd), \end{align}$$ so that $ac\theta_n^S bd$. Hence $\theta_n^S $ is compatible.    Let $i\in \mathbb{N}$ and $a\theta_i^S b$. Then for any $x=x_1\dots x_i\in S^i$, $xa=xb$, so for any $c\in S$ we have $(cx_1\dots x_i)a=(cx_1\dots x_i)b$; thus $a\theta_{i+1}^S b$. Hence $\theta_1^S\subseteq\theta_2^S\subseteq \dots$. $\square$     Lemma 2: Let $S$ be a monoid. Then for all $n\in\Bbb N$, $\theta_n^S$ is $1_S$, the identity relation on $S$.   Proof: Let $n\in \Bbb N$. If $a1_Sb$, then by definition $a=b$ and so for all $x$ in $S^n$, $xa=xb$, so that $a\theta_n^S b$ and $1_S\subseteq\theta_S^n$. If $a\theta_n^Sb$, then since $1\in S^n$ we have $a=1a=1b=b$, i.e., $a1_Sb$, so $\theta_n^S\subseteq 1_S$. Hence $\theta_n^S= 1_S$. $\square$     Lemma 3: For $n\in \Bbb N$, denote $S/\theta_n^S$ by $S_n$. Then for $n\ge 2$, $$S_n\simeq S_{n-1}/\theta_1^{S_{n-1}}.$$    Proof: User amrsa gives a proof of this (and a proof alone) in an answer below. $\square$  The Question:     Let $S=\langle a\rangle= M(m, r)$ be the monogenic semigroup generated by $a$ with index $m$ and period $r$, where $m&gt;1$. Show that   $$S/\theta_1^S\simeq M(m-1, r)$$ and deduce that   $$S/\theta_n^S\simeq M(m-n, r)$$ for $n&lt;m$.      Show also that $S/\theta_n^S$ is isomorphic to the cyclic group of order $r$ for all $n\ge m$.   My Attempt:  (The above proofs are my answers to the subproblems of Exercise 1.9.17, so I'm not approaching you empty-handed.)  I suppose $S/\theta_1^S\simeq M(m-1, r)$ because $\theta_1^S$ simply removes an $a$ from $\langle a\rangle$. I can see it but I can't prove it yet. That $S/\theta_n^S\simeq M(m-n, r)$ comes from Lemma 3 by the same token.  Any help would be appreciated.
119233
I have already tried with $S_3$, and indeed, the product is $(13)$, and $(13)^2=e$    But what about this: I define + in this way:$45=2$,$26=3$, $1$ is the identity. therefore, $123456=2326=233=3$, and as you can see from the picture, the inverse of $3$ is not itself, thus contradicts the statement.  My question: What is wrong with my argument?  And How do you prove the statement? it seems that my set of numbers perfectly forms a group under the operation I defined via the table above. first it's closed, second there is an identity, third every elements have an inverse. By the way it's commutitive. I tried (45)2=22=6=42=4(52), (45)6=26=3=43=4(56), so associativity seems to be satisfied.