Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

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.

share|cite|improve this answer

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 $\,\mathbb Z/p\mathbb{Z}$)

share|cite|improve this answer
+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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.