# If $G$ is finite and every subgroup is characteristic then $G$ is abelian and cyclic

I've been trying to prove that if $$G$$ is finite and every subgroup is characteristic then $$G$$ is cyclic. If I suppose that $$G$$ is abelian I've been able to prove it this way

The statement is true if $$|G|=1$$, so lets supose by induction that it is true for every group of order less than $$n$$.

Then if $$|G|=n=p_1^{\alpha_1} \cdots p_m^{\alpha_m}$$ with $$p_i$$ primes, as $$G$$ is abelian, we know that is direct product of its uniques Sylow subgroups

$$G=P_1P_2 \cdots P_m$$

Let $$K$$ be a subgroup of $$P_i$$ and $$f \in Aut(P_i)$$, then we can define $$h:G \longrightarrow G$$ such that given $$a \in G$$ and $$a=a_1 \cdots a_m$$ its unique expresion as a product where $$a_j \in P_j$$,

$$h(a)=h(a_1\cdots a_m)=a_1 \cdots a_{i-1} f(a_i) a_{i+1} \cdots a_m$$

The function $$h$$ is a homomorphism because given $$a, b \in G$$, if we express $$a=a_1 \cdots a_m$$, $$b=b_1 \cdots b_m$$ in the unique way as specified before, we have that

$$ab=(a_1\cdots a_m)(b_1\cdots b_m)=(a_1b_1) \cdots (a_mb_m)$$

and again, because $$a_jb_j \in P_j$$, by uniqueness of the expresion we have

$$h(ab)=(a_1b_1) \cdots (a_{i-1}b_{i-1}) f(a_ib_i) (a_{i+1}b_{i+1}) \cdots (a_mb_m)=\\=(a_1b_1) \cdots (a_{i-1}b_{i-1}) f(a_i)f(b_i) (a_{i+1}b_{i+1}) \cdots (a_mb_m)=\\=(a_1 \cdots a_{i-1} f(a_i) a_{i+1} \cdots a_m)(b_1 \cdots b_{i-1} f(b_i) b_{i+1} \cdots b_m)=\\=h(a)h(b)$$

Furthermore, $$h$$ is inyective because if $$h(a)=h(b)$$ then

$$a_1 \cdots a_{i-1} f(a_i) a_{i+1} \cdots a_m=b_1 \cdots b_{i-1} f(b_i) b_{i+1} \cdots b_m$$

and again by uniqueness of the expression of an element of $$G$$ as a product of elements of $$P_j$$ we have that $$a_j=b_j$$ if $$j=1, \cdots, i-1,i+1, \cdots m$$ and $$f(a_i)=f(b_i)$$, and as $$f$$ is injective, $$a_j=b_j$$ and $$a=b$$. We stated that $$G$$ is finite so $$h$$ is onto and we conclude that $$h \in Aut(G)$$. Now, we observe that as $$K \subset P_i$$ and it is a characteristic group of $$G$$, $$f(K)=h(K)=K$$ so $$K$$ is a characteristic group of $$P_i$$.

So, we have proven that every subgroup of $$P_i$$ is characteristic in $$P_i$$ so by induction hypothesis, every $$P_i$$ is cyclic and there exists elements $$a_1, \cdots, a_m$$ in $$G$$ such that $$o(a_i)=p_i^{\alpha_i}$$. Lastly, we observe that as the $$\gcd(p_1^{\alpha_1}, \cdots, p_m^{\alpha_m})=1$$ and $$G$$ is abelian, we conclude that

$$o(a_1\cdots a_m)=p_1^{\alpha_1} \cdots p_m^{\alpha_m}=|G|$$

and then $$G$$ is cyclic.

My doubts now are if this prove is correct and, in case it is correct, is there a way to prove in an easy way that if $$G$$ is finite and every subgroup is characteristic then $$G$$ is abelian?

Thank you very much for the comments.

• What if $G$ is a $p$-group? Then there's no smaller subgroup to use the induction hypothesis on. Commented Oct 18, 2019 at 17:21
• Note that every finite abelian group $G$ is isomorphic to $$C_{n_1}\times C_{n_2}\times \ldots \times C_{n_k}$$ for some integers $k\geq 0$ and $n_1,n_2,\ldots,n_k\geq 2$ such that $n_1\mid n_2\mid\ldots\mid n_k$. Here, $C_n$ is the cyclic group of order $n$. If $G$ is not cyclic, then $k\geq 2$. Show that $G$ has an automorphism that does not fix a subgroup isomorphic to $C_{n_1}$. Commented Oct 18, 2019 at 17:24
• @Matt Samuel You are right, I should say that since every abelian p-group is charateristically simple, in the factorization on the cardinal of G there must be at least two different primes. Do you think it would be correcto then? Commented Oct 18, 2019 at 18:40
• If the group is not abelian, then every subgroup is normal. But the finite groups for which every subgroup is normal are well known, and the nonabelian ones have a direct factor isomorphic to the quaternion group of order $8$. And that group does not satisfy the property that every subgroup is characteristic. Commented Oct 18, 2019 at 19:37
• @MattSamuel What's a gentle way of killing the fly in this particular case? I have thought a little about the problem and I don't see a direct proof. If you have one please share. Commented Oct 19, 2019 at 12:12

A partial answer: I show that $$Aut(G)$$ must be abelian, in particular $$Inn(G)$$ is abelian, and so $$G$$ is of class at most $$2$$. Perhaps one can complete the result from that point.
Let $$A=Aut(G)$$ and $$C_1,C_2,..., C_n$$ be the all cyclic subgroups of $$G$$. By hypothesis, they are all $$A$$-invariant. Let $$K_i$$ be the kernel of the action $$A$$ on $$C_i$$.
Since $$Aut(C_i)$$ are abelian we see that $$A/K_i$$ are abelian. Notice that $$K=\bigcap_{i=1}^{n}K_i=1$$
as $$K$$ must act trivially on $$G$$. Since we can embed $$A/K=A$$ to the abelian group $$\prod_{i=1}^{n} A/K_i$$ by sending $$(aK)\to (aK_1,aK_2,...,aK_n)$$, we see that $$A=Aut(G)$$ is abelian.
By using the fact that $$Aut(G)$$ is abelian, one can show that $$[G,A]\leq Z(G)$$ and $$G'\leq C_G(A)$$.