Question

If G is a group such that o(G) = 2m , where m∈N , then G has a subgroup of order m.
State the given statement is true or false, give reasons for your answer.

Answer 1

Answer on Question #78776 – Math – Abstract Algebra

Question

If $G$ is a group such that $\operatorname{ord}(G) = 2m$, where $m \in \mathbb{N}$, then $G$ has a subgroup of order $m$. State the given statement is true or false, give reasons for your answer.

Solution

1) Assume that $m = 1$. Then there are two subgroups of order 1 and 2.

2) If $m = 2$ (or 3, or 5), then by Cauchy's theorem, $G$ has an element of order 2 (or 3, or 5).

3) If $m = 4$, then by the first Sylow theorem, $G$ has a subgroup of order 4.

4) Let's consider $A_4$, the group of all even permutations of a 4-element set, i.e. products of an even number of transpositions. The order of $G$ is $\frac{4!}{2} = 12 = 2 * 6$; it consists of the identity, the one fixed points, and the double transpositions. Any subgroup of order 6 contains the identity. By Cauchy's theorem, it also contains an element of order 2 and an element of order 3, i.e. a double transposition and a fixed point, $t$ and $f$. Therefore such subgroup must contain 2 other fixed points, $tf$ and $ft$ (see the Cayley table: $tf = ft \Rightarrow (tf)^2 = (ft)^{-1} \Rightarrow ftf = f^{-1} \Rightarrow tf = e \Rightarrow |tf| \neq 3$). By the inverse element axiom, it must contain 2 other fixed points, $(ft)^{-1}$ and $(tf)^{-1}$ (this can also be directly proven). This is a contradiction $(1 + 2 + 2 + 2 > 6)$.

Answer: The statement is false. The smallest counter example is $A_4$ ($m = 6$).

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