Examples of groups that are not subgroup Sorry, my English is poor, but I have a question. It is usual to find the definition of subgroup as: "We define a subgroup $H$ of a group $G$ to be a nonempty subset $H$ of $G$ such that when the group operation of $G$ is restricted to $H$, $H$ is a group in its own right". But it is known that $A=\left\{\begin{pmatrix}a & a\\a& a\end{pmatrix} \mbox{ with  } a \in R \mbox{ and } a\neq 0 \right\}$ is a group with the usual product of matrices but it is not a  subgroup of $M_{2\times2}$. Do you know another examples of subsets like this?
Edit: I'll try to change a little the question: In the set of matrices $M_{2\times2}$ there are subsets which are groups in regard to the usual matrix product but have a different neutral element. An example would be the group of all nonsingular matrices over $\mathbb R$,  $GL(2,\mathbb R)$ and $H=\left\{\begin{pmatrix}a & a\\a& a\end{pmatrix} \mbox{ with  } a \in \mathbb R \mbox{ and } a\neq 0 \right\}$. Do you know of other examples of subsets (groups) with elements of the same nature, that given the same operation have a different neutral element?
 A: If $M$ is a monoid and $X\subseteq M$ is a subset closed under multiplication, it possible for $X$ to not contain $\mathrm{id}_M$ and yet have its own identity element, even being a group. In particular, every ring is a monoid with respect to multiplication, like matrix algebras, so we can use them.


*

*If $R\oplus S$ is a direct sum of nontrivial unital rings, then $R^\times\times\{0_S\}$ fails to contain the identity element $1_{R\oplus S}=(1_R,1_S)$, but it is a group with respect to multiplication. In particular if these are matrix algebras $R=M_n(\mathbb{F})$ and $S=M_m(\mathbb{F})$ over a field $\mathbb{F}$ then this is the group of all block diagonal matrix with $n\times n$ block an invertible matrix and all other blocks zeroed. More generally we can use $R^\times\times\{e\}$ with any idempotent $e^2=e$ in $S$.

*The ideal $(x)=\mathbb{F}x$ minus $\{0\}$ in the ring $\mathbb{F}[x]/(x^2-x)$.

*If $G$ is a group (or monoid really), we can adjoin a new identity $e$ to it, so that the multiplication table on $\{e\}\cup G$ extends that of $G$ with $eg=g=ge$ for all $g\in \{e\}\cup G$; then $G$ does not contain the identity element $e$.

A: The set of non-zero real numbers is a group under multiplication. As a subset of real numbers, it is not a subgroup under addition.
The  subset $\{-1, +1\}$ of integers is a group wrt multiplication but not a subgroup under addition of integers.
All continuous maps $f\[a,b]\to[a,b]$ that are bijective form a group under composition of functions as binary operation. It is a subset of the group of all functions $[a,b]\to \mathbf{R}$ (this being a group under point-wise addition of functions). But not a subgroup.
