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bab

▪ I. † bab Obs. A former nursery word for dad or papa. [Cf. baba; also It. babbo papa, dad.]1598 Florio, Pappa..the first word children vse, as with vs dad or daddie or bab.▪ II. bab earlier, and now dial. form of babe.▪ III. bab dial. form of bob, a bait for eels. Oxford English Dictionary

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Bab

Bab or BAB can refer to: Bab (toponymy), a component of Arabic toponyms literally meaning "gate" Set (mythology) (also known as Bab, Baba, or Seth) ancient and municipality in the Nitra District in western central Slovakia Bab Ballads, cartoons published by W. wikipedia.org

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The order of $bab^{-1}$ is equal to the order of $a$ for any $a,b \in G$ Let $G$ be a group. Prove that the order of $bab^{-1}$ is equal to the order of $a$, $\forall a,b \in G$

**Key Idea** $\ $ Isomorphisms preserve all "group-theoretic" properties, which includes the order of an element $\,g,\,$ since this equals the order (cardinality) of the cyclic group generated by $\,g.\,$ But an isomorphic image of a group has the same order (cardinality). Yours is the special case...

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Babı

Babı () is a village and municipality in the Fuzuli District of Azerbaijan. It has a population of 1,129. wikipedia.org

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Babá

Babá may refer to: Babá (footballer, born 1934), Mario Braga Gadelha, Brazilian footballer. Babá (footballer, born 1938), Santino Quarth Irmão, Brazilian footballer. Babá (footballer, born 1944), Roberto Caveanha, Brazilian footballer. wikipedia.org

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In a ring $(A,+, \cdot)$ if $aba = a$ then $bab = b$ and all non zero elements in $A$ are invertible. Let $\left(A,+, \cdot\right)$ be a ring with $1$ that satisfies the following condition: For any nonzero $a\in A$,...

Clearly then $b$ can only be $a^{-1}$ and it follows that $bab=b$ also holds.

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Babù

Anderson Rodney de Oliveira (born 23 December 1980), known as Babù, is a Brazilian football coach and former player. Playing career Early career A forward, Babù started his career off in Italy, being noted by Zdeněk Zeman in 2001 during a youth football tournament in wikipedia.org

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Prove $(bab^{-1})^n = ba^nb^{-1}$ by induction p(1): $(bab^{-1})^1 = ba^1b^{-1}.$ p(k + 1): $(bab^{-1})^{k + 1} = (bab^{-1})^k(bab^{-1}) = b^ka^kb^{-k}bab^{-1} = ba^{k + 1}b^{-1}.$ Would that work?

Then $(bab^{-1})^1 = bab^{-1}$ Assume for our induction hypothesis that $(bab^{-1})^n = ba^n b^{-1}$ for some $n\geq 1$. $(bab^{-1})^{n+1} = (bab^{-1})^n(bab^{-1}) =^{I.H.} ba^nb^{-1}bab^{-1}$ $=ba^n(b^{-1}b)ab^{-1} = ba^n 1 a b^{-1} = ba^nab^{-1}=ba^{n+1}b^{-1}$ Thus,

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Let $G$ be a finite group. If $a = bab$, is it true that $b^{2} = e$ Let $G$ be a finite group and let $a,b \in G$. If $a = bab$, is it true that $b^{2} = e$. If not, find a counterexample. It is clear that if $a = b...

No. For example, in $Q_8$ we have $jij=jk=i$.

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An interesting puzzle for some, confusing for me Suppose that $a$ is of odd order $k$ and $bab=a$. I need to show that $b$, must be of order $2$. We can prove this anyway we want to, but our hint is to expand $(bab)^k...

We have $$bab^{-1}=b^2a$$ So $$e=(bab^{-1})^k=(b^2a)^k=(b^2a)^{k-2}b^2ab^2a =(b^2a)^{k-2}a^2=\cdots =b^2a^k=b^2$$

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秋天里的春光 Babí léto

　　正所谓“少要沉稳老轻狂”，年届不惑的弗兰提斯克·汉那（Vlastimil Brodský 饰）越老越有活力，他不去理会老婆艾蜜莉（Stella Zázvorková 饰）辛苦存下棺材本钱的良苦用心，也懒得搭理那个不成器儿子糟糕的婚姻。弗兰不愿面对即将到来的死亡，只想尽情享受最后的时光。他和剧院的好友艾德（Stanislav Zindulka 饰）假扮各种身份的人，从中得到无限的乐趣，却也因此惹下不少的麻烦。垂暮的生命，因一颗青春顽皮的心而充满光芒…… 　　本片荣获2002年克里夫兰国际电影节最佳影片奖；2002年捷克金狮奖最佳女主角奖（Stella Zázvorková）、最佳剧本奖和最佳... 豆瓣

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How many pairs of digits A,B such that ABA+BAB is divisible by 74? ABA and BAB are not multiplication, they just denote digits. I realized that ABA plus BAB will result in the last 3 digits being repeated, so do I f...

ABA+BAB=111A+111B=111(A+B)=3x37(A+B) which divides 74 when A+B is even ## where A,B is an element of {1,2,3,4,5,6,7,8,9} so (A,B)=(1,1) (1,3) (1,5) (

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How to solve for $A$ in $A - BAB^T = CC^T$? Considering an unknown real symmetric matrix $A$, and two known matrix $B$ and $C$. If we have the equation: $$ A - BAB^T = CC^T $$ Can we get an analytical solution of A?

Let $vec(A)$ denote the vectorization operator, and let $\otimes$ denote the Kronecker product. We can then rewrite both sides of the equation to get $$ (I - B \otimes B) \,vec(A) = vec(CC^T) $$ Assuming invertibility, we have $$ vec(A) = (I - B \otimes B)^{-1}vec(CC^T) $$

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Can $bab$ be a generator of the following covering space? In the following covering space, can $bab$ be a generator of a covering space of $S^1\vee S^1$? Is it more natural to let $bab$ as the generator instead of $ba...

The reason is that $bab$ wraps around both circles, while $bab^{-1}$ wraps around only one circle, so altogether the generators in $\langle a, b^2, bab Note that $bab^{-1}b^2 = bab$, so both presentations are technically equivalent.

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If $A^n=0$, then $\left(BAB^{-1}\right)^n=0$ If $A^n=0$, then $\left(BAB^{-1}\right)^n=0$. (Assume everything that's necessary for the products to make sense.) _Proof_ : $$\left(BAB^{-1}\right)^n=\underbrace{\left(B...

In a slight variant on @Monadologie's answer, we can prove by induction on $n$, that $(BAB^{-1})^n=BA^nB^{-1}$, even without using $A^n=0$. (The base step $n=0$ is trivial, as is the inductive step$$(BAB^{-1})^k=BA^kB^{-1}\implies(BAB^{-1})^{k+1}=BA^kB^{-1}BAB^{-1}=BA^{k+1}B^{-1}.$$ _Then_

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bab

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bab

Bab

The order of $bab^{-1}$ is equal to the order of $a$ for any $a,b \in G$ Let $G$ be a group. Prove that the order of $bab^{-1}$ is equal to the order of $a$, $\forall a,b \in G$

Babı

Babá

In a ring $(A,+, \cdot)$ if $aba = a$ then $bab = b$ and all non zero elements in $A$ are invertible. Let $\left(A,+, \cdot\right)$ be a ring with $1$ that satisfies the following condition: For any nonzero $a\in A$,...

Babù

Prove $(bab^{-1})^n = ba^nb^{-1}$ by induction p(1): $(bab^{-1})^1 = ba^1b^{-1}.$ p(k + 1): $(bab^{-1})^{k + 1} = (bab^{-1})^k(bab^{-1}) = b^ka^kb^{-k}bab^{-1} = ba^{k + 1}b^{-1}.$ Would that work?

Let $G$ be a finite group. If $a = bab$, is it true that $b^{2} = e$ Let $G$ be a finite group and let $a,b \in G$. If $a = bab$, is it true that $b^{2} = e$. If not, find a counterexample. It is clear that if $a = b...

An interesting puzzle for some, confusing for me Suppose that $a$ is of odd order $k$ and $bab=a$. I need to show that $b$, must be of order $2$. We can prove this anyway we want to, but our hint is to expand $(bab)^k...

秋天里的春光 Babí léto

How many pairs of digits A,B such that ABA+BAB is divisible by 74? ABA and BAB are not multiplication, they just denote digits. I realized that ABA plus BAB will result in the last 3 digits being repeated, so do I f...

How to solve for $A$ in $A - BAB^T = CC^T$? Considering an unknown real symmetric matrix $A$, and two known matrix $B$ and $C$. If we have the equation: $$ A - BAB^T = CC^T $$ Can we get an analytical solution of A?

Can $bab$ be a generator of the following covering space? In the following covering space, can $bab$ be a generator of a covering space of $S^1\vee S^1$? Is it more natural to let $bab$ as the generator instead of $ba...

If $A^n=0$, then $\left(BAB^{-1}\right)^n=0$ If $A^n=0$, then $\left(BAB^{-1}\right)^n=0$. (Assume everything that's necessary for the products to make sense.) _Proof_ : $$\left(BAB^{-1}\right)^n=\underbrace{\left(B...