prove that $ [G:K]$ is finite and $ [G:K]= [G:H] [H:K]$
Well off the top of my head, by the 3rd Isomorphism Theorem $ (G/H)/ (K/H) \cong G/K$ so that would definitely take care of part 1 (just have to show why it will). Let me think about the …
If $H$ is a subgroup of $G$ and $K$ is a subgroup of $H$, then …
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If $K \leq H \leq G$, show that $ [G:K] = [G:H] [H:K]$.
Jul 5, 2023 · This is not for homework. (I am a grader for a class.) The case in which $G$ is finite is trivial. (That is, use a corollary to Lagrange's Theorem, and set $[G:H ...
abstract algebra - If $H,K < G$ and $ (G:H), (G:K)$ are finite, prove ...
Sep 15, 2021 · It would be better if instead of obliterating your previous work to which the comments refered, you simplly added the new material (with relevant added comments to …
abstract algebra - Showing that $ [G:K]= [G:H] [H:K]$ for $K \leq H ...
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Prove that $K\le G$ of finite index and $ [G:K]= [G:H] [H:K]$ for ...
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Prove that $[G:K]=[G:H][H:K] $where $K\\le H \\le G$
Feb 26, 2020 · Also this. All I did was search for the tag [group-theory] and K\leq H\leq G. It should have been easy to find lots of duplicates of this.
Let $f (z) = u+iv$ be an analytic function, then show that
Dec 13, 2018 · Real and imaginary parts of an analytic function satisfy the Laplace's equation. [$\frac {\partial ^ {2}} {\partial x^ {2}} u+ \frac {\partial ^ {2}} {\partial y^ {2 ...
Help understanding a proof that $[G:H][H:K]=[G:K]$
Oct 5, 2018 · The proposition to be proved was: If $K$ is a subgroup of $H$ and $H$ a subgroup of $G$, then $[G:H][H:K]=[G:K]$, provided these indices are finite. The proof given ...