TITLE: Not understanding the case where $G$ is abelian with every element of order $2$
QUESTION [7 upvotes]: Suppose an abelian finite group $G$ (with $o(G)>2$) has every non-identity element of order $2$. Show that there exists a non-trivial automorphism on $G$.

After a bit of searching, I found a few similar questions that have been asked, and answered on this site. This is the question whose answer has almost always been given by taking a recourse to Vector Spaces and describing some weird relationship with $\mathbb Z/2\mathbb Z$. I could not understand any one of those answers.
My background: I have done a course on Linear Algebra and therefore I know something about Vector Spaces. I have started learning Group Theory and have just read about isomorphism. I know the first law of isomorphism, but nothing more. No knowledge at all of how one suddenly talks about vector spaces while answering a question on abstract algebra. And it's funny that all the answers are quite exactly the same. Is this problem very well-known? I found it in Tipics of Algebra as a "starred" exercise.
Please explain if this question can be answered within Group Theory only. I mean, by not talking about vector spaces at all. If you are more comfortable with a vector space argument, that's fine with me, but please do explain every thing clearly. Thank you very much.

REPLY [4 votes]: Without recourse to theorems like the structure of finite abelian groups: 

You know your group is abelian, since $$e=(ab)^2=abab\implies a^{-1}b^{-1}=ba\implies ab=ba$$ 
You know that you have at least four distinct elements in your group: $e,a,b,ab$.
You know that $G\cong\{e,a,b,ab\}\times H$ for some group $H$, since $G$ is abelian.
Make your automorphism map $a\leftrightarrow b$ inside  $G\cong\{e,a,b,ab\}\times H$.