Proof of P-group Theorem (2)

proofp-groups

Theorem

For p-group $(G,\cdot )$, $p$ prime such that $|G|=p^k$. Then there exist unique integers $f_1\ge f_2\ge \dots \ge f_a\ge 1$ such that $f_1+f_2+\dots +f_a=k$, and $G\simeq {\mathbb{Z}}_p{f}_1\times \dots {\mathbb{Z}}_p{f}_a$.

Proof

Strategy: construct elements $g_1,\dots ,g_a\in G$ such that $g_1$ has the largest possible order $p^{f_1}$ among all group elements. $g_2\cdot ⟨g_1⟩$ has largest possible order $p^{f_2}$ in ${}^G{/}_{⟨g_1⟩}$. $g_3\cdot ⟨g_1,g_2⟩$ has largest order $p^{f_3}$ in ${}^G{/}_{⟨g_1,g_2⟩}$ and so on.

Show that up to twisting $g_i$ by $g_1,\dots ,gi-1$, we may assure ${\mathrm{ord}}_{}(g_i)=p^{f_i}$ in $G$ and $⟨g_1⟩,⟨g_2⟩,⟨g_3⟩,\dots ,⟨g_a⟩$ give an internal direct product decomposition

$⟹G\simeq ⟨g_1⟩\times \dots ⟨g_a⟩$ (note that $⟨g_1⟩={\mathbb{Z}}_{p^{f_1}}$, and $⟨g_a⟩={\mathbb{Z}}_{p^{f_a}}$.)

We do first two steps:

Let $g_i\in G$ have largest possible order $p^{f_1}$. $g_2\in G$ have largest possible order $p^{f_2}\mathrm{mod}⟨g_1⟩$.

If $p^{f_1}=p^k=|G|⟹G=⟨g_1⟩\simeq {\mathbb{Z}}_{p^{f_1}}$.

In this case, $a=1$, $f_1=k$.

Let $\pi :G\to {}^G{/}_{⟨g_1⟩}$ (the quotient group is commutative) $\pi (x)=x\cdot ⟨g_1⟩$

Remark:

1. ${}^G{/}_{⟨g_1⟩}$ is also a p-group. (This is by Lagrange’s Theorem)
2. $x\in G$, ${\mathrm{ord}}_G(x)=p^b⟹{\mathrm{ord}}_{{}^G{/}_{⟨g_1⟩}}(x\cdot ⟨g_1⟩)=p^{b^{\prime }}$ for some $b^{\prime }\le b$. (Note that $x\cdot ⟨g_1⟩=\pi (x)$)

${\mathrm{ord}}_{{}^G{/}_{⟨g_1⟩}}(\pi (x))\mid {\mathrm{ord}}_G(x)$

Let $p^{f_2}={\mathrm{ord}}_{{}^G{/}_{⟨g_1⟩}}(g_2\cdot ⟨g_1⟩)$ By the remark above, we have that ${\mathrm{ord}}_G(g_2)=p^c$ for some $c\ge f_2$.

Claim: $\exists i\in \mathbb{Z}$, $g_2^{\prime }=g_2\cdot g_1^i$ has order $p^{f_2}$ in $G$, and still ${\mathrm{ord}}_{{}^G{/}_{⟨g_1⟩}}(g_2^{\prime }\cdot ⟨g_1⟩)=p^{f_2}={\mathrm{ord}}_{{}^G{/}_{⟨g_1⟩}}(g_2\cdot ⟨g_1⟩)$.

In this case we will observe that $⟨g_1⟩\cap ⟨g_2^{\prime }⟩=\{e_G\}$.

Question: Where does this come from?

Examples

$G=({\mathbb{Z}}_4\times {\mathbb{Z}}_2,+)$

$g_1=(1,0)$ has order $4$, largest possible

$g_2=(1,1)\in G$ has order $4$, but $g_2+⟨g_1⟩$ has order $2$.

Remark

Group $(G,\cdot )$, with normal subgroup $N⊴G$, $g\in G$, and surjective morphism $\pi :G\to {}^G{/}_N$ $⟹{\mathrm{ord}}_{{}^G{/}_N}(\pi (g))={\mathrm{ord}}_{{}^G{/}_N}(gN)=$ smallest $n\ge 1$ such that $g^n\in N$ ($⟺g^{n}N=N$, and $(gN{)}^n=e_{{}^G{/}_N}$).

$⟨g_1⟩=\{(0,0),(1,0),(2,0),(3,0)\}$ $⟨g_2⟩=\{(0,0),(1,1),(2,0),(3,1)\}$

Thus, $⟨g_1⟩,⟨g_2⟩$ intersect non-trivially, and can therefore not give an internal direct product decomposition.

But, if we replace $g_2$ by $g_2-g_1=g_2^{\prime }$, now we get ${\mathrm{ord}}_G(g_2^{\prime })=2={\mathrm{ord}}_{{}^G{/}_{⟨g_1⟩}}(g_2^{\prime }+⟨g_1⟩)$ (note $g_2^{\prime }+⟨g_1⟩=g_2+⟨g_1⟩$)

Therefore, $⟨g_1⟩$ and $⟨g_2^{\prime }⟩$ do give an internal direct product decomposition.

Proof of $⟨g_1⟩\cap ⟨g_2^{\prime }⟩=\{e_G\}$.

Assume ${\mathrm{ord}}_{}(g_2)=c>f_2={\mathrm{ord}}_{{}^G{/}_{⟨g_1⟩}}(g_2⟨g_1⟩)$ $⟹⟨g_1⟩\cap ⟨g_2⟩=⟨g_2^{p^{f_2}}⟩$

${\mathrm{ord}}_{}({}^{p^c}{/}_{p^{f_2}})=p^{c_{0-f_2}}$

$⟹$ in $⟨g_1⟩$ this must be $⟨g_1^{f_1-(c-f_2)}⟩$

$⟹g$