Let $G$ be a group containing $24$ elements.

  1. Prove that the number of elements with order 3 is even.

  2. Prove that if $G$ is abelian, $G$ has exactly $2$ elements with order $3$.

  3. Prove that if $G$ has exactly $2$ elements of order $3$, then $G$ has a normal subgroup with index $8$.

  4. Prove that if $G$ has no elements of order 6, then $G$ is not abelian. Give an example of such a group.

For (1), I want to apply a 'grouping argument' like in the proof of Cauchy's theorem for $p=2$. Note that all elements of $G$ have either order $2$, $3$ or $4$. Elements $x\in G$ of order $3$ can be grouped into $\{x,x^2\}$ while elements $y\in G$ with order $2$ can be grouped into $\{y,y^{-1}\}$. I don't know how to proceed since there are also elements of order $4$.

For (2), I know that if $G$ is abelian, then $G\simeq C_2\times C_3\times C_4$. Since the order of an element in this direct product is the lcm of the orders in each group, writing out the elements of each cyclic group and their order, I find that the only two possibilities are $(0,1,0)$ and $(0,2,0)$.

Could someone give hints for 1, 3 and 4?

My background is an introductory course on group theory, without Sylow theory (but with group actions etc.)

  • 2
    $\begingroup$ What have you covered? Centralizers? Group actions? Sylow theory? Structure theorem of finite abelian groups? It is kinda difficult to give meaningful answers and/or hints, when we don't know for sure. $\endgroup$ – Jyrki Lahtonen May 27 '18 at 18:04
  • 1
    $\begingroup$ A few possible sources of confusion: $C_2\times C_3\times C_4$ is not the only (up to isomorphism) abelian group of order $24$. There are two others. The order of $Q_8\times Q_8\times Q_8$ is $8\cdot8\cdot8=512$. $\endgroup$ – Jyrki Lahtonen May 27 '18 at 18:06
  • $\begingroup$ @Jyrki I edited my question. We covered all the things you name, excluding Sylow theory. $\endgroup$ – Heinz Doofenschmirtz May 27 '18 at 18:08
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    $\begingroup$ For $(1)$, given any element $a$ of order $3$, consider $b = a^2$ and show that is has order $3$ as well. Note that $b^2 = a$, so we have found a 'partner' for every element of order $3$. $\endgroup$ – Badam Baplan May 27 '18 at 18:12
  • 1
    $\begingroup$ Please ask one question at a time. $\endgroup$ – Shaun May 27 '18 at 19:36

Hints: Justify the following and use them.

  1. Each element of order three generates a subgroup of order three that contains exactly two elements of order three (the generator and its inverse). Show that two such subgroups intersect trivially.
  2. If $a$ and $b$ are two elements of order three such that $b$ is not a power of $a$ show that together they generate a subgroup of order nine (you need $G$ to be abelian here).
  3. If $c$ and $c^2=c^{-1}$ are the only elements of order three show that the subgroup $\{1,c,c^2\}$ is normal.
  4. If $G$ is abelian, $a$ has order two and $b$ has order three then $ab$ has order ____. Consider a suitable symmetric group as an example.

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