Let $G$ a group of order $6$. Prove that $G \cong \Bbb Z /6 \Bbb Z$ or $G \cong S_3$. [closed]

Let $G$ a group of order $6$. Prove that:

i) $G$ contains 1 or 3 elements of order 2.
ii) $G \cong \Bbb Z /6 \Bbb Z$ or $G \cong S_3$.

I haven´t covered Sylow groups and normal groups. This is an exercise from the chapter about group actions. I have covered Lagrange and cosets.

closed as off-topic by user26857, user223391, Tom-Tom, Harish Chandra Rajpoot, user228113 Dec 17 '15 at 0:05

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• Hint : Try using the two statements in this webpage groupprops.subwiki.org/wiki/… – user10444 Mar 31 '13 at 17:30
• If it is not $\mathbb Z/6\mathbb Z$ then all elements are of order $1,2,3$. Pick an element of order $2$ and an element of order $3$ and try to deduce from there. – Thomas Andrews Mar 31 '13 at 17:32

Surely given $n\in \mathbb N, n\geq 1$, there exists a cyclic group of order $n$: In our case, there exists a cyclic group $G\cong \mathbb Z /\mathbb 6Z \cong \mathbb Z_6$.

• So that covers one option: Now, $\mathbb Z_6$ has exactly how many elements of order $2$? Use Lagrange, and determine the possible orders of subgroups. Any element of order $2$ will generate a subgroup of order $2$. $Z_6$ has only one subgroup of order $2$: What subgroup of $\mathbb Z$ has order $2$ ,and which element necessarily has order $2$?

• For any group of even order, there can exist only an odd number of elements of order $2$. Why?
We can rule out $5$ such elements for a group of order $6\;\ldots\quad$WHY?
Hence, that leaves us with $G$ having $1$ or $3$ elements of order $2$. $\mathbb Z_6$ covers one possibility. If a group $G$ of order $6$ has $3$ elements of order $2$, how does, how does this fully determine the corresponding group $G$? (For any $G\not\cong \mathbb Z_6$, all it's elements must of order $1$,$2$, or $3$.)

Now, use these facts to justify that the only groups of order $6$ must be isomorphic to $\mathbb Z_6$ or to $S_3$.

• 90intuition: feel free to follow-up with questions, if you have any, or post what conclusions you can draw from the above. – Namaste Mar 31 '13 at 18:09
• NiceAmyforthegroupi+1 – mrs Mar 31 '13 at 19:53
• I don't understand: "Surely for any group G of order n (hence a finite group), there exists a cyclic group of order n." What do you mean? – Julien Mar 31 '13 at 20:49
• @julien thanks for pointing that out. Very awkward indeed. I knew what I meant, but I didn't quite say what I meant. Fixed ;-) – Namaste Mar 31 '13 at 20:52
• Thanks I understand ! I proved the WHY's for myself. – 90intuition Apr 1 '13 at 13:47

For (ii), in the case there are three elements of order 2, let $G$ act on the set of elements of $G$ of order 2 by conjugation. Let $G \to S_3$ be the corresponding homomorphism. What can be the kernel?

If there is just one element of order $2$, all elements of $G$ must commute with that element. Consider the orders of the other elements.

• I don't understand this. So let $X=\{a,b,c \}$ where $a,b,c$ elements in G of order 2. Then $G×X → X : (g,x) ↦ g.x=gxg^{-1}$ is an action. The corresponding homomorphism is: $\varphi: G →S(G): g↦(σ_g:X→X:x↦gxg^{-1})$. If $σ_g=\text{Id}_x$ then $gxg^{-1}=x$. What can I conclude now ? – 90intuition Apr 1 '13 at 13:58

I propose a different solution (although the problem is already solved in other answers) for i, that generalises it a little bit: A group $G$ with even order has an odd number of elements of order 2. From here, i follows directly.

The proof is really simple: for an element $a\in G$, we define: $$U_a=\left\lbrace a,a^{-1}\right\rbrace$$ We have that every set $U_a$ has two elements unless $a^{-1}=a$, that only happens if $a=1$ or $a$ has order two. Now if we collect all the $U_a$ for all $a\in G$, $G$ must be the disjoint union of all of them. So the sum of all $|U_a|$ must the order of the group. We're adding a $1$ for the element $1_G$, and two for every element with order different from two, that means that the number of elements of order $2$ must be odd so they sum up to an even number

$G$ must have subgroups of orders 2 and 3 by Cauchy's theorem. These subgroups have prime order, so are cyclic, so say that $a$ and $b$ generate them and thus have orders 2 and 3, respectively.

Consider the set $\{1, a, b, ab, b^2, ab^2\}$. These must all be distinct, or else one of $a$ or $b$ will end up with the wrong order. For example, $ab = b^2$ gives $a=b$, which is impossible. Since these six products exhaust the group, $G$ is generated by $a$ and $b$.

Now, what is $ba$? It must be the same as one of the six elements above. $ba=1$ implies $a=b^{-1}$ which gives the wrong order for $a$ or $b$. Similarly $ba=a$ implies that $b$ is the identity, and has order 1 instead of order 3, and we can rule out $ba=b$ and $ba=b^2$ similarly. This leaves $ba=ab$ and $ba=ab^2$.

$ab=ba$ means that any product of $a$s and $b$s in any order can be rearranged into the form $a^ib^j$, and then to $a^{i\bmod 2}b^{j\bmod 3}$, and so is exactly the abelian group $Z_2\times Z_3 = Z_6$.

The final possibility is $ab = b^2a = b^{-1}a$. This is exactly the defining relation of $D_6$, since a reflection followed by a rotation is the same as a reverse rotation followed by a reflection. $D_6$ is isomorphic to $S_3$, with $a=(1 2)$ and $b = (1 2 3)$.

• I really like this proof! But I don't understand the first part: Why does $G$ abelian imply that $G$ cyclic ? I understand this part: $ab=ba ⇔ abb=bab ⇔ ab^2=b^2a$ . But why does this mean the group must be abelian ? – Kasper Apr 16 '13 at 14:43
• I cleaned up and shortened the proof considerably. I added details about why $ab=ba$ proves that $G$ is abelian. That an abelian group of order 6 must be $Z_6$ follows from the classification of finite abelian groups which I thought you would have studied already: Any finite abelian group has the form $Z_{n_1}\times Z_{n_2}\times\cdots\times Z_{n_k}$, and $Z_a\times Z_b\equiv Z_{ab}$ if and only if $a$ and $b$ have no common factor. So the only possible abelian groups of order 6 are $Z_2\times Z_3$ and $Z_6$, and these are the same. – MJD Apr 16 '13 at 14:58
• (Thanks very much for your questions, which addressed real problems in my explanation.) – MJD Apr 16 '13 at 14:59

If $G$ isn't $\mathbb Z/6\mathbb Z$, then it isn't cyclic. Hence, the maximal order is 3 (see Lagrange). It easily follows that an element of order 2 exists as well. From here to show that it's actually $S_3$ should be trivial. This solves the second part.

The first part follows directly from the second.

• We should take just a minute to note that $o(x)=1$ iff $x=e$ so we cannot have $O_{\text{max}}(G)=2$ – user47805 Mar 31 '13 at 17:42
• It is not useful at all to tell the OP something "...should be trivial." Nothing really is. Provide a proof, or at least some help. "It easily follows that..." Why? Again: support the claims you think are easy, since the OP wouldn't be asking the question if it was "trivial" and "followed easily". – Pedro Tamaroff Mar 31 '13 at 17:55