Short exact sequence and the order of groups

Question

Given a short exact sequence $$1 \rightarrow A \xrightarrow{\phi} B \xrightarrow{\psi} C \rightarrow 1, $$ what can I say about the relationships between the order of the groups $A, B$ and $C$? This is what I came up with: since $1 \rightarrow A \xrightarrow{\phi} B \xrightarrow{\psi} C \rightarrow 1$ is exact, $\phi$ is injective and $\psi$ is surjective. Since $\phi$ is injective, every element $x \in A$ will map to one element in $B$. This means that the order of group $A$ is lesser than or equal to the order of $B$. Since $\psi$ is surjective, there is an element $x \in B$ for every $y \in C$ with $\psi(x) = y$. This means that the order of $B$ is greater than or equal to the order of $C$. So from this follows: $$|A| \leq |B| \geq |C|,$$ but I don't know if this is right.

Answer 1

See that means $A$ is embedded in $B$ as a subgroup. So $|A|\leq |B|$ and $C$ is an image of $B$ so, $|C|\leq |B|$. Your conclusion is correct.