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?
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**Is it possible?** Yes. A simple example: $f=g$.

**What does it imply**? This is equivalent to saying that $f=\Theta(g)$. To see why:

If $f=O(g)$, there exists $c>0$ and $N \geq 0$ such that
$$
\forall n \geq N, \qquad f(n) \leq c\cdot g(n) \tag{1}
$$

If $g=O(f)$, there exists $c'>0$ and $N' \geq 0$ such that
$$
\forall n \geq N', \qquad g(n) \leq c'\cdot f(n) \tag{2}
$$

Combining (1) and (2), and choosing $M\stackrel{\rm def}{=}\max(N,N')$, $\alpha\stackrel{\rm def}{=}\frac{1}{c} > 0$, $\beta \stackrel{\rm def}{=} c'$, we get that there exist $M\geq 0$, constants $\alpha,\beta>0$ such that
$$
\forall n \geq M, \qquad \alpha f(n) \leq g(n) \leq \beta f(n) \tag{3}
$$
which by definition means $f=\Theta(g)$. This should answer both your questions.
The following are equivalent:

1. $f(n) = O(g(n))$ and $g(n) = O(f(n))$
2. $f(n) = \Omega(g(n))$ and $g(n) = \Omega(f(n))$ (using the definition from computational complexity theory, not the one from analytic number theory)
3. $f(n) = \Theta(g(n))$.