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MathematicsBourbaki Algebraic Structures Question 4.11 [Groups of exponent 2 are commutative and have order of a power of two in the finite case]
[0] [1] Beauty Is Truth
[2022-05-27 04:46:32]
[ abstract-algebra group-theory ]
[ https://math.stackexchange.com/questions/4459602/bourbaki-algebraic-structures-question-4-11-groups-of-exponent-2-are-commutativ ]

Question: If all elements of a group $\mathrm G$ other than the identity element are of order $2$, $\mathrm G$ is commutative; if $\mathrm G$ is finite, its order $\mathrm n$ is a power of $2$ (argue by induction on $\mathrm n$).

(See the answer below for an attempt.)

"Idempotent" menas $x^2=x$. In a group, the only idempotent is the identity. We say a group has "exponent $n$" if and only $x^n=e$ for each all $x\in G$. These are groups of exponent $2$. - Arturo Magidin
This is especially embarrassing because one of the first observations I made about group structure was that identities are the only idempotents given that all elements are cancellable. This seemed surprising after spending so much time on left semi-groups. - Beauty Is Truth
[0] [2022-05-27 04:46:32] Beauty Is Truth

In such a group elements $\mathrm a$ are equal to their inverses $\mathrm a'$. If $\mathrm {a,b \in G}$ then $\mathrm {ab \in G}$ and hence $\mathrm {abab = e}$. By our first observation $\mathrm {aba'b' = e}$ and hence $\mathrm G$ is commutative;

In such a group observe that the quotient with respect to any monogenous subgroup $H = \{e, a\}$ is also of exponent two given that $gHgH = ggHH = H$ for all $g \in G$. Note that this shows as a corollary that $G$ must be of even order. Suppose that $f(n)$, $n$ a positive natural number, denotes the collection of natural numbers bounded strictly between $2^n$ and $2^{(n+1)}$. If no group of exponent two with order $m \in f(n)$ exists then neither can such a group exist with order $m' \in f(n+1)$ since we have shown that this would imply the existence of a group of exponent two with order $m'/2 \in f(n)$, absurd. The base case follows from the corollary.


The problem does not ask you to prove that there are groups of exponent $2$ of order $2^n$ for every $n$ (this is trivial: just take a direct product of $n$ copies of $C_2$). It asks you to prove that if $G$ is a finite group of exponent $2$, then its order is a power of $2$. - Arturo Magidin
Initially I thought to take successive quotients by cyclic subgroups, if this process terminates in a group of nontrivial odd order then we arrive at a contradiction via Lagrange's theorem. The proof given above starts with observation that $K_4$ and $Z_2$ are the only groups of exponent 2 with order less or equal to 4, that such groups of order 5, 6, 7 cannot exist, that adjoining an element to $K_4$ yields a group of order 8, etc etc. This way of proceeding seemed very constructive which I liked. - Beauty Is Truth
You are not answering the question in the "construction". As for the problem, you can simply proceed by induction. The result is trivial if $G$ has no element of order $2$. Otherwise, take a single quotient modulo the subgroup generated by an element of order $2$. Verify the quotient has exponent $2$, and use induction to conclude $|G/N|$ has order $2^k$. Since $N$ has order $2$, it follows that $|G|=2^{k+1}$. Hand-wavy constructions of groups of order $2^n$ for all $n$ is useless and irrelevant to the problem at hand, whether you like it or not. - Arturo Magidin
What you did is both handwavy and incomplete (or at least hard to follow). You are not clear on establishing all the clauses of your induction hypothesis in the "next level". - Arturo Magidin
Nor do you verify that the quotient of a group of exponent 2 is itself of exponent 2. - Arturo Magidin
Suppose $H = \{e, a\}$ is a subgroup of $G$. $G$ is abelian so we may always form the quotient group $G/H$. The identity $gHgH = ggHH = H$ for all $g \in G$ shows that the quotient $G/H$ is of exponent two. - Beauty Is Truth
The induction needn't require these constructions. For suppose $f(1) = \{3\}$, $f(2) = \{5, 6, 7\}$, $f(3) = \{9,10,11,12,13,14,15\}$, etc. If no group of exponent two exists with order $m \in f(n)$ neither can it exist in $f(n+1)$. If we know that no group of order 3 is of exponent 2 the theorem follows. I thought the construction was of use because it seemed obvious that all groups of exponent 2 having the same order are isomorphic, constructing all finite groups of exponent 2 would entail having constructed the given group. Now that I think about it more this isomorphism isn't obvious. - Beauty Is Truth
I have no idea what you think $f$ means. It means nothing to me. It is becoming apparent that you really just want to talk to yourself in your private language and with your private rules, solving whatever problems you feel like it despite quoting one problem and claiming you are answering that rather than something else. Might I suggest you try a blog instead? - Arturo Magidin
Please be patient with me. I'm here to learn. The answer you gave which introduced the readers to Goursat's lemma was very useful. - Beauty Is Truth
"which introduced readers to Goursat's lemma"... it might have introduced you, but it is rather presumptuous to just assume that because you don't know something, neither does anyone else reading your post. Goodbye. - Arturo Magidin
Definite article is ambiguous I think. - Beauty Is Truth
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