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Let $G$ be a group. $H \leq G$ and $a \in G$, let $K= H \cap (aHa^{-1})$. Is there a bijection between $H/K$ and $aHa^{-1}/K$? Here the $H/K$ just denotes the left cosets of $K$ in $H$.

This is just my thought, so I don't konw the answer. Thank you for any help.

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If $G$ is finite, then yes because they're the same size.

If $G$ is not finite, $aHa^{-1}$ can be a proper subgroup of $H$. For example,

$$ G=\begin{bmatrix} \Bbb Q^{\times} & \Bbb Q \\ 0 & 1 \end{bmatrix}, \quad H=\begin{bmatrix} 1 & \Bbb Z \\ 0 & 1 \end{bmatrix}, \quad a=\begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix}, \quad aHa^{-1}=\begin{bmatrix} 1 & 2\Bbb Z \\ 0 & 1 \end{bmatrix}. $$

In this case, $aHa^{-1}\le H$ so $K=H\cap aHa^{-1}=aHa^{-1}$ so

$$ |H/K|=2, \quad |aHa^{-1}/K|=1 $$

and there is no bijection between them.

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