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TypoThe fundamental theorem of abelian groups says that for any finite abelian group

[tex]

A \cong \mathbb Z_{p_1^{k_1}} \oplus \ldots \oplus \mathbb Z_{p_m^{k_m}}

[/tex]

Hyperlinking breaks my raw text for some reason. I'll be using Sylow's theorems which I will refer to as theorems 1, 2 and 3 respectively. (https://en.wikipedia.org/wiki/Sylow_theorems)

In the abelian case there are [itex]\mathbb Z_6\cong\mathbb Z_2\oplus\mathbb Z_3[/itex].

Quick remark. Being a Sylow [itex]p-[/itex]subgroup is invariant with respect to conjugating, thus if [itex]H[/itex] was a unique Sylow [itex]p-[/itex]subgroup, we would have [itex]gHg^{-1} = H, g\in G,[/itex] i.e [itex]H[/itex] would be normal.

Fact.Suppose a non-abelian group [itex]G[/itex] has order [itex]pq[/itex] (primes) with [itex]q\equiv 1 \pmod{p}[/itex], then there are precisely [itex]q[/itex] Sylow [itex]p-[/itex]subgroups in [itex]G[/itex].

Proof of fact.Let [itex]n_p,n_q[/itex] be the numbers of Sylow [itex]p,q-[/itex]subgroups respectively. By assumption [itex]q>p[/itex]. By theorem 3 [itex]n_q \equiv 1 \pmod{q}[/itex] and by theorem 1 [itex]n_q\mid p[/itex], which forces [itex]n_q = 1[/itex], thus we have a (normal) subgroup of order [itex]q[/itex], which makes it cyclic, therefore abelian.

For [itex]n_p[/itex] we thus have two choices: [itex]n_p = 1,q[/itex]. If [itex]n_p = 1[/itex], then [itex]G\cong \mathbb Z_p \oplus \mathbb Z_q[/itex] would be abelian, thus it must be that [itex]n_p = q[/itex].

Our non-abelian group [itex]G[/itex] therefore contains three Sylow [itex]2-[/itex]subgroups, call them [itex]H_1,H_2,H_3[/itex], which by theorem 2 are all conjugate to each other. We have [itex]\mbox{Sym}(H_1,H_2,H_3) \cong S_3[/itex]. We show that [itex]G\cong S_3[/itex]. Define

[tex]

\varphi : G\to \mbox{Sym}(H_1,H_2,H_3), \quad\varphi (g)(H_j) := gH_jg^{-1}

[/tex]

(also called conjugating)

To show it's an isomorphism, it suffices to show it's injective (because [itex]|G| = |S_3| = 6[/itex]). Suppose [itex]\varphi (g) = \mbox{id}[/itex] i.e [itex]gH_jg^{-1}=H_j[/itex], then by definition [itex]g\in N(H_j)[/itex] the normaliser of [itex]H_j[/itex]. If [itex]N(H_j) = G[/itex], then [itex]H_j[/itex] would be normal, which would then make it equal to its conjugates, contradicting the fact that there are three Sylow [itex]2-[/itex]subgroups.Firstly, if [itex]H[/itex] is a Sylow [itex]p-[/itex]subgroup, then conjugating it gives another Sylow [itex]p-[/itex]subgroup, hence it must hold that [itex]gH_ig^{-1} = H_j[/itex] for some [itex]j[/itex].

The map [itex]\varphi (g)[/itex] is injective, because if [itex]gH_ig^{-1} = gH_jg^{-1}[/itex], then

[tex]

h_i \in H_i \Rightarrow gh_ig^{-1} \in gH_ig^{-1} = gH_jg^{-1} \Rightarrow h_i \in H_j.

[/tex]

The argument is symmetrical, thus [itex]H_i=H_j[/itex]. It is surjective due to finiteness.

Take [itex]g,h\in G[/itex], then

[tex]

\varphi (gh) (H_j) = (gh)H_j(gh)^{-1} = g(hH_jh^{-1})g^{-1} = g (\varphi (h)(H_j)) g^{-1} = \varphi (g) (\varphi (h)(H_j)) = (\varphi (g) \varphi (h))(H_j).

[/tex]

Therefore, all [itex]N(H_j) = H_j[/itex] and [itex]g\in H_1\cap H_2\cap H_3 = \{e\}[/itex] i.e [itex]g=e[/itex], which makes [itex]\varphi[/itex] an isomorphism.

By theorem 3 [itex]n_q \equiv 1 \pmod{q}[/itex] and by theorem

**3**[itex]n_q\mid p[/itex][tex]

\mathbb Z_8 \qquad \mathbb Z_4 \oplus \mathbb Z _2 \qquad \mathbb Z_2 \oplus\mathbb Z_2\oplus\mathbb Z_2

[/tex]

Suppose [itex]G[/itex] is non-abelian of order [itex]8[/itex]. If there is an element of order 8, then [itex]G[/itex] is cyclic, thus abelian. Suppose the maximal order is [itex]2[/itex] and pick [itex]g,h\in G[/itex]. If [itex]gh=e[/itex], then they commute. Suppose [itex](gh)^2 = e[/itex], then

[tex]

gh = geh = g(ghgh)h = (gg)hg(hh) = ehge = hg

[/tex]

and again [itex]G[/itex] would be abelian. Thus we must have an element [itex]a[/itex] of order [itex]4[/itex]. Its generated subgroup [itex]\langle a \rangle =: H[/itex] is of index [itex]2[/itex], therefore normal. Take [itex]b\notin H[/itex], then we have [itex]bab^{-1}\in H[/itex] due to normality. Notice that

[tex]

(bab^{-1})^4 = bab^{-1}bab^{-1}bab^{-1}bab^{-1} = baaaab^{-1} = e.

[/tex]

If also [itex](bab^{-1}) ^2 = e[/itex], then [itex]baab^{-1} = e[/itex] would lead to [itex]aa=e[/itex], contradicting the order of [itex]a[/itex].

Next, we find the possible values of [itex]bab^{-1}[/itex].

- If [itex]bab^{-1} = a^4 = e[/itex], then [itex]a=e[/itex], which is impossible.
- If [itex]bab^{-1} = a[/itex], then [itex]ba = ab[/itex], but this will make [itex]G[/itex] abelian. Indeed, pick [itex]g,h\in G[/itex]. If they're both in [itex]H[/itex], then they commute. Suppose [itex]g\notin H[/itex] and write [itex]g = ba^m[/itex] and [itex]h = a^n[/itex], then

[tex]

gh = ba^ma^n = ba^na^m = a^n ba^m = hg.

[/tex]

and if both reside outside [itex]H[/itex], then writing [itex]g = ba^m[/itex] and [itex]h = ba^n[/itex] would similarly lead to [itex]gh=hg[/itex]. But our group is not abelian. - If [itex]bab^{-1} = a^2[/itex], then [itex](bab^{-1})^2 = e[/itex] would contradict the order of [itex]bab^{-1}[/itex].

There are two cases to consider.

- [itex]b[/itex] is of order [itex]2[/itex], then we have [itex]G \cong \langle a,b \mid \mbox{ord}(a) = 4, \mbox{ord}(b) = 2, bab^{-1} = a^{-1} \rangle [/itex] this is the dihedral group [itex]D_8[/itex].
- [itex]b[/itex] is of order [itex]4[/itex]. We show [itex]G[/itex] is the dicyclic group [itex]\mbox{Dic}_2[/itex], where

[tex]

\mbox{Dic}_2 \cong \langle x,y \mid x^{4} = e, x^2 = y^2, x^{-1} = y^{-1}xy \rangle.

[/tex]

Notice that

[tex]

bab^{-1} = a^{-1} \Rightarrow b^{-1}ab = b^{-1}(ba^{-1}b^{-1})b = a^{-1}.

[/tex]

So we have left to show [itex]a^2 = b^2[/itex]. Considering the natural projection [itex]G\to G/H\cong\mathbb Z_2[/itex], we have [itex]b^2 \in H[/itex]. Consider possible values of [itex]b^2[/itex].- [itex]b^2 = a^4 = e[/itex] contradicts order of [itex]b[/itex].
- [itex]b^2 = a[/itex] contradicts order of [itex]a[/itex]
- [itex]b^2 = a^2[/itex] is what we want.
- [itex]b^2 = a^{-1}[/itex] contradicts order of [itex]a^{-1}[/itex] (hence order of [itex]a[/itex]).

Thus [itex]b^2 = a^2[/itex] must hold.

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