# $g\in G$ maximal order in $G$ abelian then $G=\left<g\right>\oplus H$

If $g\in G$ is an element of maximal order in a finite abelian group $G$ then exists $H\leq G$ such that $G=\left<g\right>\oplus H$

Attempt: Using fundamental theorem I know that $G=C_{n_1}\times\cdots\times C_{n_k}$ where $n_i|n_{i+1}$. With some work I proved $|\left<g\right>|=n_k$, so $C_{n_k}\cong \left<g\right>$. Then $G\cong H\times \left<g\right>$, but this is not an equality.

• This is actually used to prove the fundamental theorem, so I wouldn't use it. – Pedro Tamaroff Apr 6 '14 at 2:01
• Oh, teacher's proof uses it :(. – Gaston Burrull Apr 6 '14 at 2:16
• possible duplicate of Finite abelian groups - direct sum of cyclic subgroup – mez May 1 '14 at 8:59
• @mezhang I didn't know how to do the $p$-group case, so this is not a duplicate. – Gaston Burrull May 1 '14 at 22:54

We know any finite abelian group is a sum of $p$ primary abelian groups (it's Sylow subgroups), so you may assume that $A$ is a finite abelian $p$-group.

Let $x$ be an element of maximal order. Note in this case $|x|=\exp A$. We can assume $A$ is not cyclic (why?), and that $\exp A>p$, else $A$ is a vector space over $\Bbb F_p$ and nonzero elements belong to a basis, and the claim is trivial.

Since $A$ is not cyclic, there exist more than one subgroup of order $p$: since $\langle x\rangle$ is cyclic, it contains a subgroup of order $p$. Let $K$ be a subgroup of order $p$; not contained in $\langle x\rangle$. Then $K\cap \langle x\rangle=0$: $K\cap \langle x\rangle$ is a proper subgroup of $K$ (why?), so it must be zero. It follows that $$\frac{\langle x\rangle+ K}K\simeq \langle x\rangle$$ is cyclic in $G/K$.

Note that $g+K$ in $G/K$ has order diving $|g|\leqslant |x|$, so $\dfrac{\langle x\rangle+ K}K$ is a cyclic subgroup in $G/K$ of maximal order, i.e it is generated by an element of maximal order in $G/K$. By induction on the order of the group, there exists a direct summand $H/K$ for some $K\leqslant H\leqslant G$. It follows that $G=\langle x\rangle +K+H=\langle x\rangle +H$. But $(\langle x\rangle+K)\cap H=K$ implies $H\cap \langle x\rangle=0$, for $K\leqslant H$. So $G=\langle x\rangle \oplus H$.

I will let you finish the proof by using $A=\bigoplus_p A(p)$ where the sum runs over the prime divisors of $|A|$. You'll have to prove that the the element in the $p$-primary parts of $A$ give rise to one element in all of $A$ of maximal order (Hint: sum them).

As I commented, this gives a partial result towards the fundamental theorem: take any abelian group and pick an element of maximal order. Then $A=\langle x\rangle\oplus H$. By induction on the order of the group, we may write $H=C_{n_{t-1}}\oplus \cdots C_{n_1}$ with $n_1\mid n_2\mid \cdots \mid n_{t-1}$. Since $n_t=|x|=\exp A$, $n_{t-1}\mid n_t$ and $A=C_{n_t}\oplus C_{n_{t-1}}\oplus \cdots C_{n_1}$ is a sum of cyclic groups with $n_1\mid n_2\mid \cdots \mid n_{t-1}\mid n_t$.

• Why I can assume $p$-group?, I dont understand – Gaston Burrull Apr 6 '14 at 4:03
• I understand now, I worked in details in your proof without difficulty. Ty. – Gaston Burrull Apr 7 '14 at 3:52

Here is a proof using the structure theorem. By taking a primary decomposition of $G$ we can assume that there is a prime $p$ and exponents $a_i \in \mathbb{N}$ such that $n_i = p^{a_i}$ for each $i$. Let

$$G = \langle g_1 \rangle \oplus \ldots \oplus \langle g_k \rangle$$

where $g_i$ has order $p^{a_i}$ for each $i$ and $a_1 \le \ldots \le a_k$. Let

$$g = c_1g_1 + \cdots + c_{k-1}g_{k-1} + c_kg_k$$

where $c_i \in \mathbb{N}$ for each $i$.

If every $c_ig_i$ has order dividing $p^{a_k-1}$ then $g$ has order dividing $p^{a_k-1} = n_k/p_k$, a contradiction. Therefore there exists $i$ such that $a_i = a_k$ and $p$ does not divide $c_i$. By reordering the $g_i$, we may assume that $i = k$. Take $r$ such that $rc_k \equiv 1$ mod $p^{a_k}$. Then $rg = (rc_1)g_1 + \cdots + (rc_{k-1})g_{k-1} + g_k$ and it is straightforward to show that $G = \langle g_1, \ldots, g_{k-1} \rangle \oplus \langle g \rangle.$

Remark. The reduction to the primary case is needed to make this argument work. For example, if $G = \langle g_1 \rangle \oplus \langle g_2 \rangle$ where $g_1, g_2$ both have order $6$ then $2g_1 + 3g_2$ is an element of order $6$, but neither $2g_1$ nor $3g_2$ has maximal order.

• I can follow your response unless "we can assume that there is a prime $p$..." What about $G=C_2\times C_{3^2}$? there are distinct primes here. – Gaston Burrull Apr 6 '14 at 4:24
• That's where we need the primary decomposition: suppose that the distinct primes dividing $|G|$ are $p_1, \ldots, p_k$. Let $G = G_{p_1} \oplus \cdots \oplus G_{p_k}$ be the primary decomposition of $G$, where $G_{p_i}$ is a $p_i$-group for each $i$. Let $g = x_1 + \cdots + x_k$ where $x_i \in G_{p_i}$. My answer gives a complement for $\langle x_i \rangle$ in the group $G_{p_i}$ for each $i$. The direct sum of these complements is a complement for $g$ in $G$. (In your example $G$ is cyclic, so $\langle g \rangle = G$ for any $g$ of maximum order.) – Mark Wildon Apr 6 '14 at 4:53
• Im almost sure that $x_i$ has maximal order in $G_{p_i}$. Is true that order of $g$ is the product of all $x_i$ orders? in that case I am right. For complements do you mean that $<g>=<x_1>\oplus \ldots \oplus <x_k>$? – Gaston Burrull Apr 6 '14 at 5:44
• Yes the order of $g$ is the product of the orders of the $x_i$: more generally, if $G = H \times K$ and $g = (h,k) \in G$ then the order of $g$ is the lowest common multiple of the orders of $h$ and $k$. For complements: for each $i$, take a complement $H_i$ in $G_{p_i}$ to $\langle x_i \rangle$. Then $H_1 \oplus \cdots \oplus H_k$ is a complement in $G$ to $\langle g \rangle = \langle x_1 \rangle \oplus \cdots \langle x_k \rangle$. – Mark Wildon Apr 6 '14 at 10:33