# Is my reasoning behind my construction of my semi direct product correct?

Classify up to isomorphism the groups of order $203$. Assume the face that the least $k \geq 1$ such that

$$2^k \equiv 1 \mod{29}$$

is $k = 28$. [HINT: Look at the Sylow subgroups and represent a group $G$ of order $203$ as a semi-direct product of two cyclic groups.]

First thing I do, is split $203 = 7 \times 29$. So now I want to work out $n_7, n_{29}$, which is the number of subgroups of order 7 and 29.

From the hint, I know that the lowest $k \geq 1 : 2^k \equiv 1 \mod 29 = 28$. We can see that $2^{28} > 203$ and so there is only 1 subgroup of order 29, i.e $n_{29} = 1$. As there is only one subgroup, this group is clearly normal.

Now, for my $n_7$, I want the number where I get $n_7 \equiv 1 \mod 7$ and $n_7 | 29$. Listing all numbers that are congruent to $1 \mod 7$, I get 2 possibilities for $n_7: n_y = 1 \, \mathrm{or}\, 29$.

When $n_7 = 1$, we get that this is normal and as we now have 2 normal subgroups, we can see that the only SDP is the trivial one (as their intersection is just the identity element), i.e the direct product such that

$$G \cong C_{7} \times C_{29} \cong C_{203}.$$

Let $C_{7} = \langle a | a^7 = 1 \rangle, C_{29} = \langle b | b^{29} = 1 \rangle$ and $\mathrm{Aut}(C_{29}) = \langle \mu | \mu^{28} = 1 \rangle$. Now, when $n_7 = 29$, I want to create a homomorphism

$$\theta :C_7 \rightarrow \mathrm{Aut}(C_{29}) \cong C_{28}.$$ $$\mu : b \rightarrow b^2$$

I still don't get why it's $b^2$ as I thought $\mu$ was supposed to map $b \rightarrow b^k$ where $k$ is the lowest integer such that $k^i \equiv 1 \mod m$, where $i \in \mathbb{Z}$, $m$ is the order of the group $\mathrm{Aut}(C_n) \cong C_m$. And this $k$ isn't 2 as we need gcd($k,m$) = 1 by Eulers theorem and gcd($2,28$) = 1. Wait, unless it supposed to be $k^i \equiv 1 \mod n$, where $n$ is the order of the group $C_n$ (before the automorphism)?

Any way, carrying on, now I know that I want $\theta: C_7 \rightarrow C_{28}$, I want to find elements of order $7$ in $C_{28}$. I can then raise $\mu$ to this power and so this will give me my $\theta(a^{-1})(b)$ which will let me get my multiplication law. So I am looking for some $k$ such that $a^k = 7$

$$7 = \frac{28}{(k, 28)}$$

Rearranging this gives

$$(k, 28) = 4.$$

Double checking: $(4, 28) = 4$, we get one element of order $7$ in $C_{28}$ to be $\mu^4$.

So, we know get our $\mu$ mapping

$\mu^4 : b \rightarrow b^{2^4} = b^{16}$

From here, we can now define our multiplication law as

$$ba = a\theta(a^{-1})(b)$$

which goes to

$$ba = ab^{16}.$$

I think that's the final answer.

Is this correct? Also, if I do one more question on SDP's, would I be able to post a similar question up for someone to check and make sure I have got everything correct?

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$$\tiny{\text{please don't yell...}}$$ – apnorton Jan 12 '13 at 1:34
There is a sandbox. Doesn't that work? – Martin Jan 12 '13 at 1:36
Didn't you already asked this question one or two days ago? About the hint with that power of two I've no idea how that works easier than directly applying Sylow theorems...In fact, I think that messes things up seriously. – DonAntonio Jan 12 '13 at 1:50
@DonAntonio Yeah I did but then I hardly had a clue what I was doing. Now I've worked through the question with a bit more knowledge so I was hoping I've done the right things for the right reasons. – Kaish Jan 12 '13 at 1:54
For example: $$(b^i,a^j)(b^r,a^k):=(b^i\cdot(b^r)^{a^j},a^ja^k)$$ where $$(b^r)^{a^j}=b^{16rj}$$ since $\,b^a:=b^{16}\,$ . Note that we use the shortwritting $\,b^a:=\mu(b):=b^{16}\,$ , to avoid the messy $\,\theta(a)(b)=\mu(b)=b^{16}\,$ . You can read about this in Rotman's book or in Robinson's, for example. – DonAntonio Jan 12 '13 at 2:32

The Sylow subgroups of $G$ have the form $C_7=\langle a \rangle$ and $C_{28}=\langle b \rangle$. Let $\alpha$ be a generator of $\text{Aut}(C_{29})$. Then $\alpha$ has order $28$, so $\alpha^4$ has order $7$. Thus we can embed $C_7\rightarrow \text{Aut}(C_{29})$ by $\theta:a\mapsto \alpha^4$.
Now, what is $\alpha$? $2$ has multiplicative order $28$ modulo $29$, so we can pick $\alpha:x\mapsto x^2$. Then $\alpha^4:x\mapsto x^{16}$. From here, the method to obtain the presentation is given in the comments.
You also asked in the comments about $\langle a,b|\text{stuff}\rangle$. This is a group presentation. They work like this: $$\langle \underbrace{a_1,a_2,\ldots,a_n}_{generators}|\underbrace{\overbrace{a_1^2,a_2^3,\ldots, a_n^7}^{\text{orders of }a_i},\overbrace{a_1^{-1}a_3a_2=a_4^2,a_2a_1=a_3a_1}^{desired relations}}_{relators}\rangle$$
Oh ok, thanks. This bit explains the whole $x \mapsto x^2$ a lot better than I do. Thank you very much! So am I looking for the number which has multiplicative order $-1 \mod 29$ or $1 \mod 29$, or neither, it is supposed to be $28 \mod 29$, which just happens to be $-1 \mod 29$? – Kaish Jan 12 '13 at 11:22
You are looking for something which has multiplicative order $7$ mod $29$. I'll explain in your newer thread. – Alexander Gruber Jan 12 '13 at 17:02