# Every nonabelian group of order 6 has a non-normal subgroup of order 2 (revisited)

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 $$x≠y, |x|=|y|=3$$. $$$$ and $$$$ are normal in $$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=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.

• 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.' – Berci Jan 4 '19 at 12:05
• @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. – Frank De Geeter Jan 4 '19 at 19:40

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.
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 $$G\cong D_3\cong S_3$$.