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I have some trouble with groups. Say we know that $A$ is a subgroup of $B$. If we have some subset of $A$, say $H$, can we deduce that $H$ is also a subgroup of $B$?

Thank you.

So if I have set of $2\times 2$ real matrices of form $\left( \begin{array}{cc} -a & b \\ b&a\end{array} \right)$ I only know its subset of $SL_2(\mathbb{R})$ but I have to show it's actually subgroup of $SL_2(\mathbb{R})$.

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  • $\begingroup$ You have to check that $H$ satisfies the group axioms. $\endgroup$ – N. F. Taussig Apr 26 '15 at 20:09
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No, that's not true. In order to be a subgroup, it's necessary (but not sufficient) for a subset $H$ to contain the identity element (that is, the one usually denoted $0$, $1$, or $e$). So you can take any group $B$ you want, and then take any subgroup $A\subseteq B$ you want, and the subset $H=A\setminus\{e\}$ cannot be a subgroup of $B$.

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No we cannot, for example if a subset consisting of exactly one element is never a subgroup, unless that element is $e$.

On the other hand if $H$ is a subgroup of $A$ and $A$ is a subgroup of $B$ then $H$ is a subgroup of $B$.

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$\newcommand{\GL}{\mathrm{GL}}\newcommand{\SL}{\mathrm{SL}}\newcommand{\R}{\mathbb{R}}$In the particular case at hand, the set $$ H = \left\{ \begin{bmatrix} -a & b\\b & a\end{bmatrix} : \text{$a, b$ not both zero} \right\} $$ is not a subgroup because the determinant of $\begin{bmatrix} -a & b\\b & a\end{bmatrix}$ is $- (a^2 + b^2) < 0$, so the product of two such matrices has positive determinant, and is not in $H$.

You probably mean $$ K = \left\{ \begin{bmatrix} a & -b\\b & a\end{bmatrix} : \text{$a, b$ not both zero} \right\}, $$ and this is a subgroup, as $$ \begin{bmatrix} a & -b\\b & a\end{bmatrix} \cdot \begin{bmatrix} c & -d\\d & c\end{bmatrix} = \begin{bmatrix} a c - b d & - (a d + b c) \\a d + b c & a c - b d\end{bmatrix} \in K, $$ and $$ \begin{bmatrix} a & -b\\b & a\end{bmatrix}^{-1} = \frac{1}{a^2+ b^2} \begin{bmatrix} a & b\\-b & a\end{bmatrix} \in K. $$

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  • $\begingroup$ Yes, I meant something like that. Or for example, matrices like those in your K but with a^2+b^2=1. Such set K would be subset in SL_2(R). Am I right? $\endgroup$ – user222801 Apr 27 '15 at 5:46
  • $\begingroup$ Yes, same proof. $\endgroup$ – Andreas Caranti Apr 27 '15 at 6:02

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