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I am fully aware that this question has already been addressed here and here. The question, however, derives from a Dummit and Foote exercise (section 4.2 exercise 10 page 122) and the answers provided make use of material that appears later in the book, or, at least, I do not clearly understand them in terms of the material that I have studied already.

So, I would like to submit the following tentative proof 'from first principles' (i.e. Dummit and Foote before page 122). As I feel insecure about it, I would be grateful if you could check it.

Consider a nonabelian group G of order 6. By Cauchy's theorem, the group contains at least one element of order 2, and therefore at least one subgroup of order 2. Suppose this/these subgroups are all normal. This would imply that all elements of order 2 commute with all elements of G.

The remaining elements of group G are of order 1 (which trivially is in the center of G), or order 3 (order 6 would imply the group is cyclic and therefore abelian). Since all elements of order 2 are in the center, the order 3 elements will commute with them. The following reasoning shows that they also commute with each other.

Let $x≠y, |x|=|y|=3$. $<x>$ and $<y>$ are normal in $G$ because their index is $2$. Now $x\{1, y, y^2\}=\{1, y, y^2\}x$ implies, in the nonabelian case, that $x.y=y^2.x$ and $x.y^2=y.x$. Likewise, $y<x>=<x>y$ implies $y.x=x^2.y$ and $y.x^2=x.y$. Manipulating these equations shows $y.x = x.y$, so the nonabelian case is impossible. Hence, order 3 elements are in the center of G as well.

Therefore, G would be abelian, contrary to the assumption. So, at least one of the subgroups of order 2 should be non-normal.

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    $\begingroup$ Can you justify this sentence: 'Since all elements of order 2 are in the center, the order 3 elements will be also in the center.' $\endgroup$ – Berci Jan 4 '19 at 12:05
  • $\begingroup$ @Berci. Spot on! I overlooked the fact that order 3 elements also would need to commute with each other. However, I might have a patch for that, which I edited into the post for further checking. $\endgroup$ – Frank De Geeter Jan 4 '19 at 19:40
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Basically your reasoning is correct (well done!): to write it with somewhat more "sophistication" - if $N \lhd G$, with $|N|=2$, then $N \subseteq Z(G)$, that is what you are using. Now if $x \in G$ with $ord(x)=3$, then, since $N$ is central, $N \subsetneq C_G(x)$, where the inclusion is strict, because of $|N|=2$ not divisible by $3$ ($x \in C_G(x)$). But $|G:N|=3$, so $G=C_G(x)$, meaning $x \in Z(G)$ and hence $G$ is abelian.

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Once you have Cauchy's theorem available it is clear that any non-abelian group $G$ of order $6$ is isomorphic to $S_3$ -- which has a non-normal subgroup of order $2$. Indeed, by Cauchy's theorem there exist elements $r$ and $s$ in $G$ of order $3$ and $2$ respectively. The subgroup $C_3=\langle r\rangle$ is normal in $G$ because it is of index $2$. So $G=C_3\cup C_3\cdot s$ and $G=\{e,r,r^2,s,rs,r^2s\}$. Because $C_3$ is normal, $srs^{-1}=r^k$ for some $k=0,1,2$. Since $s^2=e$, we have $$ r=s^2rs^{-2}=s(srs^{-1})s^{-1}=r^{k^2} $$ so that $k^2\equiv 1 \bmod 3$, i.e., $3\mid (k-1)(k+1)$. For $3\mid (k-1)$ the group is abelian, so that we have $3\mid (k+1)$ and $G\cong D_3\cong S_3$.

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