Not understanding the case where $G$ is abelian with every element of order $2$

Question

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.

Answer 1

It can be easily proven using the structure theorem that finite abelian groups in which every element has order $2$ or $1$ are of the form $\underbrace{\mathbb Z_2\times \mathbb Z_2\dots \times \mathbb Z_2}_n$.

One non trivial automorphism for a group of this form sends $(a_1,a_2\dots ,a_n)$ to $(a_2,a_1\dots ,a_n)$

Answer 2

The fundamental theorem on abelian groups (namely it's a product of cyclic groups each of prime power order) means your group is a product of 2 or more copies of ${\mathbb{Z}_2}$ (otherwise you could find an element of higher order than 2). Say there are $n$ copies. Then the group of automorphisms contains a subgroup (maybe even the whole group) isomorphic to $S_n$, i.e. the group of permutations of your $n$ copies of ${\mathbb Z}_2$. In other words, the element with a single 1 in one of its $n$ entries and 0 in all other entries can be mapped to any other such element, so you can permute the positions of the 1's any way you want.

Answer 3

The reason for the automatic switch of language to "vector spaces mod 2" is that it is one of two equivalent descriptions of what any such group looks like (up to isomorphism), but is has the advantage over the second description (which writes the group as a product of 2-element subgroups) of being canonical. The decomposition of such a group into independent 2-element pieces requires additional arbitrary choices, exactly like choosing a basis in a vector space. For purposes of theory and understanding, it is usually better to avoid thinking in basis-dependent or noncanonical terms.

Also, once the vector space structure is recognized, all the facts of linear algebra (over finite fields) can come into play.

The vector structure is not so bad: elements of the group are the vectors, and the group multiplication is addition. Identity element is the $0$. There is no need to worry about scalar multiplication because the only scalars here are $0$ and $1$.

Since you asked about automorphisms, here there is a big advantage of the vector space point of view. A construction of all automorphisms, and a means of analyzing the structure of the automorphism group by analogy with ordinary linear algebra over real, complex or rational numbers rather than finite fields. To construct the automorphisms, choose a basis (this does not contradict what I wrote earlier!) and write down any $n \times n$ matrix filled with $0$'s and $1's$ such that the determinant $\mod 2$ is equal to $1$. The linear transformations represented, in the chosen basis, by such matrices are all automorphisms of the group. With or without choosing a basis, the structural analysis of the automorphisms can proceed as in the case of a real vector space. For example, there is a unique automorphism moving any $k$ linearly independent points to any other $k$ linearly independent points, where $k$ is the dimension of $G$ as a vector space, because this is a property of linear maps. None of this is nearly as visible in the "finite abelian groups" point of view.

Answer 4

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 $G\cong\{e,a,b,ab\}\times H$.