Characteristic subgroups of order $2$ Could anybody give an example of a finite abelian $2$-group with more than one characteristic subgroup of order $2$ ? (In other words, a finite abelian $2$-group $G$ with a Klein subgroup $V$ such that every element of $V$ is fixed by every automorphism of $G$.) Thanks in advance.
 A: $\newcommand{\Span}[1]{\langle #1 \rangle}$$\newcommand{\Order}[1]{\lvert #1 \rvert}$There is no such example.
Suppose $$G = \Span{x_{1}} \times \dots \times \Span{x_{n}},$$ where, setting $\Order{x_{i}} = 2^{e_{i}}$, we have $e_{1} \ge e_{2} \ge \dots \ge e_{n} > 0$.
Setting $y_{i} = x_{i}^{2^{e_{i} - 1}}$, we have that the involutions of $G$ are the non-trivial elements of $$\Omega(G) = \Span{y_{1}, \dots, y_{n}}.$$
Now consider the automorphism $\beta_{i}$, for $i > 1$, that fixes all $x_{j}$, except that it maps $x_{i}$ to $x_{i} x_{i-1}^{2^{e_{i-1} - e_{i}}}$. Thus $\beta_{i}$ maps $y_{i}$ to $y_{i} y_{i-1}$, for $i > 1$.
Now every subgroup of order $2$ except $\Span{y_{1}}$ can be written as  $\Span{y_{i} z}$, for some $i > 1$, and some $z \in  \Span{y_{1}, \dots, y_{i-1}}$. Now $\beta_{i}$ maps $y_{i} z$ to $y_{i} y_{i-1} z \ne y_{i} z$. 
It follows that the only characteristic subgroup of order $2$ could be $\Span{y_{1}}$, and then only if $n = 1$, or $e_{1} > e_{2}$. The case $n = 1$ is clear. If $n > 1$ and  $e_{1} = e_{2}$, then the automorphism that  interchanges $x_{1}$ and $x_{2}$, and fixes all $x_{i}$ for $i > 2$,   also interchanges $y_{1}$ and $y_{2}$. On the other hand, if $n > 1$ and $e_{1} > e_{2}$, then 
$$
\Span{y_{1}} = G^{2^{e_{1} -1}}
$$
is characteristic in $G$.
Actually, there is nothing special about the prime $2$ here, the same argument works to show that a finite abelian $p$-group has at most one characteristic subgroup of order $p$.
