If $H$ has the same (finite) order as $G$, then any surjective map is necessarily injective. 
If $H$ has the same (finite) order as $G$, then any surjective map is necessarily injective.

From Dummit and Foote 3 ed., page 39.
A line in the text and not sure why it is true. If more information is needed, I can edit my post and add the more of the paragraph, or link the post to a pdf copy of the textbook.
 A: Injectivity and surjectivity of a map from a finite set to itself are equivalent conditions. The fact that it's a group doesn't contribute anything, it's just a basic fact about finite sets.
A: If $H$ and $G$ are finite groups, then the sets $H$ and $G$ are also finite. If there exists a surjection $f:H\to G$, then that means that every element in $G$ is mapped to by $f$. i.e. $\forall g \in G, \exists h \in H$ s.t. $f(h) = g$. 
So if $f$ is surjective, why is $f$ necessarily injective? Because $f$ is only not injective if there are two or more elements in $H$ that map to the same element of $G$, i.e. if $f$ is not injective, then $\exists h, h' \in H$ s.t. $f(h) = f(h')$ but $h \neq h'$. However, if $H$ is surjectively mapped onto $G$, and $|G| = |H|$, then we don't have any further elements remaining in $H$ that we can map onto $G$. This is really just an application of the pigeon-hole principle if you think about it. 
A: Let the cardinalities of the two groups $G,H$ be equal and finite and let $\varphi:H \rightarrow G$.  If $\varphi$ is not surjective, then by the Pigeonhole principle, there is some $g \in G$ which is the image of more than one element of $H$, making $\varphi$ not injective.  If $\varphi$ is not injective, then by the Pigeonhole principle, there is at least one element of $G$ that is not the image of any element of $H$, making $\varphi$ not surjective.  Therefore $\varphi$ injective $\iff$ $\varphi$ surjective.
