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Let $G$ be an finite abelian group, define $s= \Pi_{g\in G} g$, show that $s = e$ unless $G$ has exactly one involution $u$, in this case $s=u$.

I have the start of a proof, but I get stuck at a point.

Proof

Write $G$ as $G = \{e\} \cup I \cup \{g_1, g_1^{-1}, \ldots, g_n, g_n^{-1}\}$ where $I$ is a set of the all the groups involutions.

Since $G$ is abelian $\Pi_{g\in \{g_1, g_1^{-1}, \ldots\}} g = e$ we just need to examine $\Pi_{u\in I} u$

Now:

  • If $|I|=0$ then the theorem holds,
  • If $|I| = 1$ then $I=\{u\}$ and $s= u$, the theorem holds

  • If $|I| = 2$, requires closer examination:

    Let $I = \{u_1,u_2\}$ be the set of all involutions (not equal to $e$), then $u_1u_2$ is also an involution (since $G$ is abelian).

    If $u_1u_2 =e$ then $u_1=u_2$, contradiction, if $u_1u_2 = u_1$ then $u_2=e$, contradiction. $u_1 u_2=u_2$ results in the same contradiction. This means that the case $|I|=2$ is not possible.

  • the case $|I|=3$ is possible, let $I = \{u_1,u_2,u_3\}$ then it's easy to prove $u_3 = u_1u_2$. The theorem holds.

How should I continue? I've tried splitting $I$ as

$$I=\{u_i: u_i \not = u_ju_k (\text{for a certain }j\not = k)\} \cup \{u_i = u_ju_k (\text{for a certain} j \not = k)\}$$ but I still get stuck.

Also the case $|I|= 4$ seems contradictory once again..

dietervdf
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    The set of all involutions, together with $e$ - that is, the set of all $g$ such that $g^2 = e$ - is a subgroup. Show that its order is a power of $2$. – Daniel Fischer Aug 20 '17 at 20:05

1 Answers1

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Let $I = \{ u_1 , \cdots, u_n \}$ and assume $n > 2$. Note that $I$ is a subgroup.

Now we construct a set $S_n$ by the following inductive procedure: Let $S_1 = \{ u_1 \}$ and $S_2 = S_1 \cup \{ u_2 \}$. Now let $k \in \{3, \cdots, n\}$. If $u_k$ can be written as a product of elements in $S_{k-1}$, let $S_k := S_{k-1}$. Otherwise, $S_k := S_{k-1} \cup \{ u_k \}$.

Thus, $S_n$ is a generating set of $I$. (I'm renaming it $S$ for now on.) Let $S = \{ s_1, \cdots, s_m \}$. A key fact about $S$ is that it guarantees unique representations. Suppose $I_1, I_2 \subseteq \{ 1, \cdots, m \}$ are two nonempty sets of indicies such that

$$ \prod_{i \in I_1} s_i = \prod_{j \in I_2} s_j $$

or, because the $s_i$ are involutions,

$$ \prod_{i \in I_1} s_i \prod_{j \in I_2} s_j = e.$$

By the construction of $S$, this implies $I_1 = I_2$.

Thus, every index set $I \subseteq \{ 1, \cdots, m \}$ corresponds to a unique element $\prod_{i \in I} s_i$. So $2^m = n$.

Now we can complete the proof. You already covered the cases $|I| = 1$ and $|I| = 2$.

If $|I| > 2$, then $|I| = 2^m$ for some $m \geq 2$. Let $S = \{ s_1, \cdots, s_m \}$ be the generating set of $I$ again. Consider the product

$$ \prod_{u \in I} u = \prod_{I \subseteq \{1, \cdots, m \} } (\prod_{i\in I} s_i ). $$

On the right hand side, each $s_i$ appears $2^{m-1}$ times. This is because there are $2^m$ subsets of $\{ 1, \cdots, m\}$, and exactly half contain $s_i$. Since $m \geq 2$, $2^{m-1}$ is even. So the $s_i$ terms all cancel out, and the product is $e$.

aras
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