# Elements of order $2$ in an abelian group generating a non-cyclic subgroup

If $G$ is Abelian group and $a,b \in G$ are distinct elements of order $2$, show that $ab$ has order $2$. Prove that $H=\{e,a,b,ab\}$ forms a subgroup of $G$ that is not cyclic.

I request help on how I can show that $ab$ has order $2$ as well as to show $H$ is not cyclic. I think there exist no $x \in G$ that generates $H$.

Thanks all.

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$(ab)(ab) = aabb = ee = e$ to show $(ab)$ has order $2$. And to show that $H$ is not cyclic $a$ and $b$ are distinct and $a^2 = b^2 = e$ so $a^n \neq b$ for any $n$ and similarly $b^k \neq a$ for any $k$. – Deven Ware Dec 17 '11 at 21:02
@DimitrijeKostic $a^{-1} = a$ since they have order 2 ... (so $b\neq a^{-1}$ as they are assumed distinct) – Deven Ware Dec 17 '11 at 21:03
$G$ abelian implies $(ab)^2 = a^2 b^2$. – user17762 Dec 17 '11 at 21:03

At this level, there are many things you have to show. Let's do it in small steps.

1. First show that $ab$ does not have order $1$, i.e. that $ab \neq e$. To do this, assume that $ab = e$ and derive a contradiction to the assumption that $a \neq b$. Remember that $b^2 = e$.
2. Now you want to show that $(ab)^2 = e$. Write out $(ab)^2 = abab$ and remember that the group is commutative, so in particular $a$ and $b$ commute.
3. Show that $H$ is actually a subgroup. It contains the identity element of $G$, so that's good. Write out a multiplication table to prove that $H$ is closed under the group operation. This will also show that $H$ contains inverses for all of its elements.
4. Show that $H$ has order $4$. We know that neither $a$ nor $b$ is equal to $e$ because $e$ has order $1$, and we assumed that $a\neq b$, so $\{e, a, b\}$ has three elements. You showed in (1) that $ab \neq e$, and now show that $ab = a$ and $ab = b$ both lead to contradictions, and hence that $\{e, a, b, ab\}$ has $4$ elements.
5. Check that each element of $H$ generates a subgroup of $H$ of order less than $4$. Remember what this means: an element $x$ of a group generates a cyclic subgroup consisting of all the powers of $x$: $\ldots, x^{-2}, x^{-1}, e, x, x^2, \ldots$ If $x$ has finite order $n$, then there will only be finitely many elements in this list, and in fact $\{e, x, \ldots, x^{n - 1}\}$ is the complete list of distinct elements of the subgroup generated by $x$. If you haven't proved this, then you should! Thus $a$ generates a subgroup of order $2$, and so on.

It's also fun to prove that the cyclic group of order $4$ and this group $H$ (which is known as the Klein four-group) are the only groups of order $4$ up to isomorphism.

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i think e generates {e},a generates{e,a},b generates {e,b} and ab generates {e,ab} all have order less than 4.thus H is not cyclic – neema Dec 18 '11 at 3:35
@neema Looks good to me! – Dylan Moreland Dec 18 '11 at 3:37
will i be wrong to say each element in H is an inverse of itself? – neema Dec 18 '11 at 3:40
@neema That sounds fine as well. For each $x \in H$ you have $xx = e$, so $x = x^{-1}$. – Dylan Moreland Dec 18 '11 at 3:42
@neema I think it should be $aa = e$ (the diagonal entries will all be $e$), but your $a(ab)$ looks good. – Dylan Moreland Dec 18 '11 at 3:50

Try writing out what $(ab)^2$ is and using the fact that you're in an abelian group. Then, ask yourself, what does it mean for a group to be cyclic? It means that there's an element with order equal to the order of the group, right? If you can show that there is no such element, then you've shown the group is not cyclic.

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$(ab)^2 = (ab)(ab)= a(ba)b$ (Associativity)

$=a(ab)b$ (since $ab=ba$, since $G$ is Abelian),

$(ab)^2 = (a^2)*(b^2)= e\cdot e,$ since $|a|=|b|=2$,

$(ab)^2 = e,$ so $|ab| \leq 2,$ but if $|ab| = 1,$ then $a = inv(b) = b,$ which is a contradiction, since $a$ and $b$ are distinct. Thus, $|ab| = 2.$

$$\forall x \in H, |x| \leq 2 < 4, \quad \therefore x \neq H, \forall x \in H.$$

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