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Suppose $G$ is finite and abelian. Show that every subgroup of $G$ is characteristic if and only if $G$ is cyclic.

I have the 'if' part so far:

If $G$ is cyclic, then $G = \langle g \rangle $ with $|g| = n$, say. Let $\alpha \in \operatorname{Aut}(G)$, then $\alpha : G \rightarrow G$ with $g \mapsto g^i$ where $(i,n) = 1$. Let $K \leq G$, then $K$ is cyclic with $K = \langle g^k \rangle$ for some $k \in \mathbb{Z}$. $$\alpha (K) = \alpha ( \langle g^k\rangle ) = \langle \alpha(g)^k \rangle = \langle (g^i)^k \rangle = \langle (g^k)^i \rangle =: H.$$ Obviously $H\leq K$. Pick an arbritrary element in $K$, $(g^k)^j$ say $(g^j) = (g^i)^{j'}$ for some $j'$ as $g^i$ generates $G$. So $$(g^k)^j = (g^j)^k = ((g^i)^{j'})^k = ((g^k)^i)^{j'} \in H.$$ hence $K\leq H$ and hence $H=K$, thus every subgroup is characteristic.

But the 'only if' part gives me trouble. this is what I've come up with:
$G$ is finite abelian, hence $$G = C_{a_1}\times C_{a_2} \times \cdots \times C_{a_m}$$ where each $C_{a_i}$ is cyclic and $a_{i} \mid a_{i+1}$ (correct me if I'm wrong but I think this is called Smith normal form?). Suppose $m\geq 2$. From here I'm trying to find an automorphism of $M:=C_{a_1} \times C_{a_2}$ which does not fix every subgroup of $M$, but unfortunately I've had no luck.

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2 Answers

Hint: If $1<a_1$ and $a_1\mid a_2$, then $(g_1,1)\mapsto (g_1,1)$, $(1,g_2)\mapsto (g_1,g_2)$ extends to an automorphism of $C_{a_1}\times C_{a_2}$. Here $g_i,i=1,2,$ are the respective generators.

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Well, following your own stuff, suppose $\,G\,$ is not cyclic. Prove then that there exists a prime $\,p\,$ s.t. $\,C_p\times C_p\,$ is a subgroup of $\,G\,$, and show that a non-cyclic elementary $\,p\,$ group can never fulfill the condition that all its subgroups are characteristic (hint: an elementary abelian $\,p-$group is a vector space over $\,\mathbb Z/p\mathbb{Z}$)

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+1 This is how I started thinking about it. You left the task of lifting the automorphism to all of $G$ as an exercise for the reader :-) –  Jyrki Lahtonen Jun 2 '12 at 19:41
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