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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> 
> Can two function be Big-O of each other?
> 
> 
> 

The answer is yes.

One may take $f(n)=n$ and $g(n)=2n+1$, as $ n \to \infty$, one has
$$
\left|\frac{f(n)}{g(n)}\right|=\frac{n}{2n+1}\le \frac12 \implies f=O(g)
$$ and
$$
\left|\frac{g(n)}{f(n)}\right|=\frac{2n+1}{n}\le 3 \implies g=O(f).
$$
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. $f(n) = \Theta(g(n))$.