order of subgroup same as order of group(finite groups) If I have order of a subgroup C of same order as group G I want to prove that G = C.
One inclusion is obvious C $\subset$ G the other inclusion we can get by a bijection 
f : G $\rightarrow$ C hence for $x \in G$ we have $f(x) = y \in C$ and we can have a bijection since we have a bijection between G and C.
What do you guys think of this argument does it looks good ?
 A: No this is not a proof. After all if it would work, then it would also work for infinite groups. A correct proof is: the complement of $C$ in $G$ has cardinality equal to the cardinality of $G$ minus the one of $C$, which is $0$. 
Or: the inclusion $C\rightarrow G$ is always injective, and in this case also bijective (because it is an injection of finite sets of the same cardinality).
A: Although the question if formulated in terms of group theory, the group structure plays no role at all: this is really a set-theoretic question. The condition that $G$ is finite is important, because otherwise the statement would be false (For example, the group $\Bbb Z$ has proper subgroups which are in bijection with $\Bbb Z$). 
You can solve it by beginning, as you did, with a bijection $f:G\to C$, which exists since $G$ and $C$ are assumed to have the same order. If you consider the restriction $f|_C: C \to C$, it must be injective (as the restriction of an injective map $f$), hence surjective (since an injective map from a finite set to itself is always surjective). If there existed an element $x\in G$ with $x\notin C$, then by the surjectivity of $f|_C$, we could find an element $c\in C$ with $f(c)=f(x)$, contradicting that $f$ is one-to-one.
