# Whether there exists or not, a group of $2$ elements $a$ and $b$ with $\lvert b \rvert = 2 , \lvert a\rvert = 15$ and …

Does there exist a group with two elements $a$ and $b$ with $\lvert b\rvert = 2, \lvert a\rvert = 15$ and $bab = a^4$?

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What is the norm $\lvert \cdot \lvert$ in your problem? Are we talking about subgroup of the real numbers? –  Thomas Aug 30 '12 at 14:01
I guess that he is talking about the order of the element. –  Sigur Aug 30 '12 at 14:04
@Sigur: That would make sense. –  Thomas Aug 30 '12 at 14:05
I know these non abelian groups of order $30$ $$\mathbb Z_3\times D_{10}=<x^2=1,aba=b^4>$$ $$\mathbb Z_5 \times D_6=<x^2=1,aba=b^{-4}>$$ $$D_{30}=<x^2=b^{15}=(ab)^2=1>$$ –  B. S. Aug 30 '12 at 14:13
You should provide more context/ more of your background to get answers that are useful for you. Such a group is easy to construct, if you know about semidirect products and automorphism groups of cyclic groups. Is this a homework problem? –  ladisch Aug 30 '12 at 14:27

Yes. There is such a group. It is a semi-direct product of cyclic groups of respective orders 15 and 2. The reason why this works is that the mapping $x\mapsto x^4$ is an automorphism of order 2 of the cyclic group generated by $a$. Therefore we have a homomorphism from $C_2$ to $\mathrm{Aut}(C_{15})$, and can build the semi-direct product in the usual way.

If you want a concrete version of such a group, here it is as a subgroup of $S_{15}$. Let $a$ be the 15-cycle (here $A=10,B=11,\ldots$) $$a=(123456789ABCDEF),$$ so $a^4=(159D26AE37BF48C)$. Then let $b$ be the product of six disjoint 2-cycles $$b=(25)(39)(4D)(7A)(8E)(CF).$$ The usual conjugation trick of permutations then shows that $$bab^{-1}=bab=a^4.$$ The group has 30 elements. They are either of the form $a^i$ or of the form $ba^i$, $0\le i<15$.

To make sure. We don't have too many choices for the integer $m$ in the relation $bab^{-1}=a^m$. This is because it has to fit together with the fact that $b^2=1$, so $b^2$ has to commute with $a$. The following calculation uses this $$a=1a1=b^2ab^{-2}=b(bab^{-1})b^{-1}=ba^mb^{-1}=(bab^{-1})^m=(a^m)^m=a^{m^2}.$$ For this to hold, we must have $m^2\equiv1\pmod{15}$. This leaves are four choices (pairwise non-congruent modulo $15$): $m=1$, $m=-1$, $m=4$ and $m=-4$. Of these, the first choice leads to the direct product $C_{15}\times C_2$, the choice $m=-1$ gives us the dihedral group $D_{15}$ of 30 elements. The third choice gives the group in question.

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Thanks so much so your help. I'm a newcomer in group theory, so I know there are so many things I must learn:-D –  le duc quang Aug 30 '12 at 14:49
@leducquang: You are welcome! As an extra exercise I invite you to show that the element $ab$ is of order six. Do it in two ways: by computing the cycle structure, and by using the relation as described in mixedmath's helpful answer. You will learn about the use of relations from the latter way, so if you only have time for one, do it that way! –  Jyrki Lahtonen Aug 30 '12 at 15:05

You could also manually give the presentation $G = \langle a, b | a^{15} = b^2 = 1, bab = a^4 \rangle$

We might ask, how big is this group? I claim that it's not so hard to see that this group has at least $30$ elements. Does it have more?

Note that $bab = a^4 \iff a = ba^4 b$ as $b^{-1} = b$. So given any word, we can use this to simplify the word quite a bit. In particular, we need to know how to simplify words that look like $ba^3ba^2ba^5$, some alternating set of $b$ and powers of $a$. But an $a$ will be sandwiched between two $b$ terms in such a word. In this case, I might take the $a^2$ term and write it as $(ba^4b)(ba^4b) = ba^8b$. In general, $a^n = ba^{4n}b$.

Thus $ba^3ba^2ba^5 = b(ba^{12}b)ba^2ba^5 = a^{14}ba^5= a^{14}b(ba^{20}b) = a^{14}a^{20}b = a^4b$, for example.

More directly, we might also want to see that $ba^j = b(ba^{4j}b) = a^{4j}b$, so that we can write any two term word in the form $a^n b$. Combining these ideas together (more formally that I do here), we see that the only words that are left are $\{1, a, a^2 , \ldots a^{14}, b, ab, a^2b, \ldots, a^{14}b\}$, and so this is a group of order $30$. (In fact, I believe this is the same group as in the other answer, but with a much less clever presentation).

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The group generated by $a, b$ with $a$ of order 15, $b$ of order 2, and $bab=a^4$ is a group of order 30. Any element can be written as a power of $a$ followed by a power of $b$.
The subgroup $H$ generated by $a^3$ is of order 5 and normal: $ba^3b=(bab)^3=(a^4)^3=a^{-3}$.
The subgroup $K$ generated by $a^5$ and $b$ is abelian of order 6: $ba^5b=(a^4)^5=a^5$
Also $G=HK$ but the group is not the direct product. The center is cyclic of order 3 generated by $a^5$.