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whence
whence, adv., conj. (n.) (hwɛns) Forms: 4–5 whannes, whennes, (4 huannes, wannes, whennus, -ys), 4–6 whens, 5 qwens, 6 whense, Sc. quhens, quhence, 6– whence. [13th c. ME. whannes, whennes, f. whanne, whenne + -s suffix1. In all senses often preceded by redundant from, † fro (from 15 a), occas. of (...
Oxford English Dictionary
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From Whence Came the Cowboy
From Whence Came the Cowboy is the fifth album from the Sons of the San Joaquin and the third and final for the Warner Western label.
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en.wikipedia.org
How to Use Whence vs. from whence Correctly - GRAMMARIST
Whence, according to its conventional definition, means from where, so the phrase from whence is logically redundant. But this doesn't stop people from using from whence, a phrase that has been common for centuries. When hearing the sentence Whence came you? , one may feel something is missing—specifically, a preposition—even though the ...
grammarist.com
Is it Wrong to Say 'From Whence'? - Merriam-Webster
The fact is that both the phrase and the bare adverb have been used for centuries, and there is nothing wrong with either. Whatever the condemnations that sometimes are made, from whence is well established, and you should feel free to use it, or not. We won't think you vicious either way. The word means 'from what place, source, or cause'.
www.merriam-webster.com
whence
whence/wens; ?@ hwens; hwɛns/ adv(arch or fml 古或文) from where 从该处; 从那里 They have returned whence they came. 他们从哪儿来的已经回哪儿去了.
牛津英汉双解词典
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What is meant by "A and B represent the same functor whence are isomorphic" in this solution? While browsing some old questions I came across the following: tensor product of sheaves commutes with inverse image It se...
Let $\mathcal{C}$ be a (locally) small category and let $F \colon \mathcal{C} \to \bf{Set}\hspace{1mm}$ be a functor, then we say that $F$ is representable if there is a natural isomorphism between $F$ and $\textrm{Hom}_{\mathcal{C}}(A, -)$ for some object $A \in \mathcal{C}$. In this case, we say t...
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Let $G$ be a finite group and $\phi:G \to K$ be a surjective homomorphism and $n \in \mathbb{N}. $ If $K$ has an element of order $n,$ so does $G.$ Let $G$ be a finite group$\ ,\phi:G \to K$ be a surjective homomorphi...
The proof is OK. But you might use the well known and generally useful fact that a cyclic group of order $m$ has a cyclic subgroup (and hence elements) of any order $d$ dividing $m$, to reduce this to a one-liner. With $o(w)=n$, $\phi(g)=w$ and $m=o(g)$ one has $e_K=\phi(e_G)=\phi(g^m)=w^m$, so $n$ ...
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How to compute cyclic notation (23)(1) I seem to become confused whence computing simple cyclic notations as such. From my understanding, the rule goes by starting from the right and to the left. However by doing thi...
$(23)(1) = (1)(23)$ since the cycles are _disjoint_ , and disjoint cycles commute. $(23)(1)$ means $1$ maps to $1$, $2$ maps to $3$, and $3$ maps to $2$. When an element maps to itself, we can omit such a one-cycle, and it is implicitly understood that the missing element maps to itself. So, the mos...
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The Basis for the Legitimacy of the Chinese Political System: Whence ...
Philip C. C. Huang has been teaching at the Renmin University of China since his retirement from the University of California, Los Angeles, in 2004. He has now completed the third volumes of both his study of Chinese civil justice (published in both English and Chinese) and his study of Chinese agriculture (forthcoming from Falü chubanshe, in Chinese only) from the Ming-Qing down to the present.
journals.sagepub.com
Whence legitimate peripheral participation? De-trivializing the social ...
Indeed, according to the Confucian Analects, "when a disciple enters, he is filial (弟子入则孝)," i.e., students need to be careful about what they say and what they do, should not contradict the teacher. Meanwhile, "the Master, by orderly method, skillfully leads men on (夫子循循然善诱人)."
www.sciencedirect.com
If $M$ is either Noetherian or Artinian and $M^{(n)}\cong M^{(m)}$, then $m=n$. The question is: Prove that if $_RM$ is either artinian or noetherian and if $m,n\in\mathbb{N}$ with $M^{(m)}\cong M^{(n)}$, then $m=n$....
If $M^{(m)} \cong M^{(n)}$ via the map $g$ and we assume $n<m$, then if $f: M^{(n)} \to M^{(m)}$ is an inclusion map, we have $g \circ f$ is an monomorphism of $M^{(n)}$. But then if $M$ and therefore $M^{(n)}$ is Artinian, this means that $g \circ f$ is an automorphism. But this is impossible becau...
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All the eigenvalues of $A^*A$ are non-negative ? Let $A$ be a $m\times n$ matrix with complex entries and let $A^*$ be it's conjugate transpose , then off-course $A^*A$ is a Hermitian matrix whence all its eigenvalues...
Let's see: if $\lambda$ is an eigenvalue of $A^*A$ then for some nonzero $x$, $$ \lambda \|x\|^2 = \langle \lambda x,x \rangle = \langle A^*A x,x\rangle = \langle Ax,Ax \rangle \ge 0.$$
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If $N$ and $M$ are normal subgroups and $N$ and $M$ have no common element other than $e$ then prove that for all $m \in M$ and $n\in N$, $mn=nm$. My approach ; I proved the $MN$ to be a normal sub group whence $mn=nm$.
Try showing $n^{-1}m^{-1}nm\in N\cap M$.
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If $G$ is a group and $[G:Z(G)]=4$, show that $G/Z(G)$ is isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$ > I want to show that if $G$ is a group and $[G:Z(G)]=4$, then $G/Z(G)$ is isomorphic to $\mathbb{Z}_2 \times ...
I am guessing that you mean you wish to prove that $G/Z(G) \cong \mathbb{Z}_2 \times \mathbb{Z}_2$, because otherwise your statement implies that $|G| = 16$. Assuming this, begin by showing that if $G/Z(G)$ is cyclic, then the group $G$ must be abelian. Hence, since there is only one non-cyclic grou...
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