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Can you please check my logic and maybe suggest some other strategies for the following problem:

Prove that every group of order $4$ is abelian as follows: Let $G$ be any group of order $ 4$, i.e., $|G| = 4$.

  • (1) Suppose there exists $a \in G$ such that $o(a) = 4$. Prove that $G$ is abelian.

  • (2) Suppose that no element of $G$ has order 4. Prove that $\forall x\in G$, $x^2 = 1$.

  • (3) Suppose that no element of $G$ has order 4. Prove that $G$ is abelian.

What I got so far:

  • (1) If there exists $a \in G$ such that $o(a) = 4$,

    • Case 1: $a\cdot a=b$. Then $a\cdot a\cdot a=c$ and $a\cdot a\cdot a\cdot a=1$. Algebra... G is abelian.

    • Case 2: $a\cdot a=c$... $G$ is abelian.

  • (2) Let $x\in G$. If $o(x) \neq 4$, we can clarify that an element cannot have an order greater than $4$ in a group of order $4$ and that the only element that has an order of $1$ is $1$. Therefore the other three elements must have an order of $2$, so $x^2=2$ for all $x \in G$.

  • (3) No ideas yet :(

Shaun
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John Long
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  • Notice that the elements of a group of order must be e,a,b,ab or e,a,b,ba where e is the identity element. – Andrew Shedlock Oct 30 '19 at 01:45
  • In case 1, you don't need two sub-cases. The names $b$ and $c$ are arbitrary, so the second proof is exactly the same, except that you switched up the names of the elements. But the names don't matter. This is also why you don't need separate proofs for each of $a,b,$ or $c$ being the element of order $4$. You decided to call the element of order $4$ by the name “$a$”, that's fine; in the same way you can also decide to call the element $a\cdot a$ by the name “$b$”. – MJD Oct 30 '19 at 11:26

4 Answers4

3

Another strategy to 1)

If there exist some $a \in G$ such that $O(a) = 4$ then $G = \{e,a,a^{2},a^{3} \}$

Thus $c,d \in G \implies c= a^{i}, d = a^{j} \implies c\cdot d = a^{i} \cdot a^{j} = a^{i+j} = a^{j} \cdot a^{i} = d\cdot c$

3) Suppose there no exist $a \in G$ such that $O(a) = 4$

Then if $a \in G, a \not= e \implies O(a) = 2$

Let $a,b \in G$ then $a \cdot b \in G \implies (a\cdot b)^{2} = e \implies (a\cdot b)(a\cdot b) = e \implies a\cdot b \cdot a \cdot b = e $

Thus $a \cdot (a\cdot b \cdot a \cdot b) \cdot b = a \cdot e \cdot b = a \cdot b \implies a^{2}\cdot b \cdot a \cdot b^{2} = a\cdot b \implies e \cdot b \cdot a \cdot e = a \cdot b$

$ \implies b \cdot a = a \cdot b $

Then $G$ is an abelian group.

ZAF
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  • Thanks a lot! What do you think about (2)? Is it ok? – John Long Oct 30 '19 at 02:25
  • About (2), I can say this: $o(a) $ divides $ |G|$ for all $a \in G$ then $o(a) = 1,2$ or $4$, then as only $e$ have order $1$, if $a \not = e \implies o(a)$ must be $2$ – ZAF Oct 30 '19 at 02:28
1

1)

If $o(a) = 4$ then the elements of $G$ are $a,a^2, a^3, e$ and so for for every $b,c \in G$ there are unique $j,k$ so so that $b=a^j, c=a^k$ and so by associativity $bc = a^ja^k =(a*..*a)(a*...*a)=(a*.....*a) =a^{j+k} = (a*..*a)*(a*...*a) = a^k*a^j = cb$.

3)

Let $(ab)^2 = abab = e$.

$ab = a(e)b = a(abab)b = (aa)ba(bb) =e(ba)e = ba$

fleablood
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To get yourself familiar with small groups and working with elements, I would suggest you write out the multiplication table for your group! If the table is symmetric, then $G$ must be abelian. Say that $G$ has four elements, so $G = \{1, a, b, c\}$ where $1$ is the identity element.

Let’s try $(1)$ using your notation. I know the following facts:

  1. $a = a$
  2. $a \cdot a = b$
  3. $a\cdot a\cdot a = c$
  4. $a\cdot a\cdot a\cdot a = 1$

so my multiplication table looks like

$$\begin{array}{|c|c|c|c|c|} \hline~ & 1 & a & b & c \\ \hline 1 & 1 & a & b & c \\ a & a & b & c & 1 \\ b & b & c & \cdot & \cdot\\ c & c & 1 & \cdot & \cdot \\ \hline\end{array}$$

where you should fill in the rest! Convince yourself that I don’t need to do the case where $a\cdot a = c$ — the phrase “without loss of generality” applies here.

Your proof for $(2)$ looks good, so let’s look at $(3)$. Now, we know three things:

  1. $a\cdot a = 1$
  2. $b\cdot b = 1$
  3. $c\cdot c = 1$

so our table looks like

$$\begin{array}{|c|c|c|c|c|} \hline~ & 1 & a & b & c \\ \hline 1 & 1 & a & b & c \\ a & a & 1 & \cdot & \cdot \\ b & b & \cdot & 1 & \cdot\\ c & c & \cdot & \cdot & 1 \\ \hline\end{array}$$

To fill this in, you should pick an entry you don’t have an element for, and ask yourself what could make sense to put there. For example, we don’t have an entry for what $a \cdot b$ is, yet. Consider the cases

  • $a\cdot b = 1$
  • $a\cdot b = a$
  • $a\cdot b = b$
  • $a\cdot b = c$

and convince yourself that $a\cdot b = c$ is the only option. Then, try this for the other entries, and see what your table looks like!

Santana Afton
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(1): Suppose $G$ has an element $a$ of order $4$, so $|a|=|G|$. It follows that $a$ generates $G$ and thus $G$ is cyclic. Since $G$ is cyclic, we know $G$ is abelian.

(2): Suppose $G$ does not have any elements of order $4$. Consider $a \in G$. By Lagrange's theorem, we know that $|a| \mid |G|$, so $$ |a| \in \{1,2,4\}. $$ By assumption, $|a| \neq 4$, so $|a|=1$ or $|a|=2$. In either case, we have $a^2=1$.

(3): Suppose $G$ does not have any elements of order $4$. We claim that $G$ is abelian. Let $a,b \in G$. From part (2), we know $a^2=b^2=(ab)^2=1$, so $a=a^{-1}$, $b=b^{-1}$, and $ab=(ab)^{-1}$. It follows that $$ ab=(ab)^{-1}=b^{-1}a^{-1}=ba. $$ Since $ab=ba$, $G$ is abelian.

This is only a specific case of a more general theorem. More generally, any group of order $p^2$ where $p$ is prime is abelian. See this answer for more info.

Dastur
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