2 votes

Inverse bijections between subgroups of $G$ containing the kernel and subgroups of $H$ contained in the image.

It isn't true that $(\varphi^{-1} \circ \varphi)(A)=A$. Suppose $G = \mathbb{Z},\ A = \{0\},\ H = \{0\}$ and $\varphi(n) = 0\quad \forall n \in \mathbb{Z}$ then $(\varphi^{-1} \circ \varphi)(A) = G \...
1 vote

Proving equivalence of two definitions of "field"

I'm going to try to clarify your question before i give the answer to my interpretation of it... A field (F,+,x) is often either defined using 9 axioms or by simply saying: $(F,+)$ and $(F\setminus\{...
  • 100

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