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I am working on a problem that asks to classify groups of order 33. I have the following work listed below as a first attempt, however I'm aware of another method that uses a homomorphism into an automorphism group of order 11. This other method should yield the same answer I have below I believe. See below:

Let $|G|$ = 33. Then 33 = 3*11, which implies we have at least one sylow 3 subgroup and at least one sylow 11 subgroup. The number of sylow 3 subgroups divides 11, so it is 1 or 11. The number must also equal 1 mod 3, which implies we can only have 1 sylow 3 subgroup as 11 mod 3 = 2. Similarly we can conclude we have one sylow 11 subgroup.

We call $H$ our sylow 3-subgroup and $K$ our sylow 11 subgroup. Then $H$$\times$$K$ = $G$ as both $H$ and $K$ are normal and $H$$K$ = $G$.

I believe the above is correct, however I am interested in a method that uses a homomorphism into the automorphism group of $K$, I am not sure how to use this idea but I am aware that such a proof technique exists. I'm hoping someone knows how to do this and can fill in the details for me.

Thanks!

H_1317
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1 Answers1

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Hint The automorphism group of the cyclic group $C_n$ is $C_n^\times$, and if $q$ is prime, then $$\#C_q^\times = q - 1 .$$

So, for $p, q$ prime, if a homomorphism $\phi : C_p \to \operatorname{Aut}(C_q) \cong C_q^\times$ is nontrivial, we must have $p \mid (q - 1)$.

Remark Essentially the same argument shows that the only group of order $pq$, where $p, q$ are primes (and w.l.o.g. $p < q$) satisfying $p \nmid (q - 1)$, is $C_p \times C_q \cong C_{pq}$.

Travis Willse
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  • How are we defining the homomorphism and why must n| q-1? So 3 divides 10 implies a contradiction in our specific case, why can we conclude the group must classify as C_3 cross C_11? – H_1317 Oct 31 '19 at 15:49
  • I should have guessed that this is a duplicate, even with the restriction on the method of proof. See this excellent answer for example: https://math.stackexchange.com/a/1502364/155629 If we have subgroups $H, N \leq G$ such that $H \cap N = {1}$ and $N$ normal, the characterization of semidirect products tells us that $G \cong N \rightthreetimes_\phi H$ for some homomorphism $\phi : H \to \operatorname{Aut}(N)$. – Travis Willse Oct 31 '19 at 18:48
  • If $H = C_p$, $N = C_q$, for a generator $\bar 1 \in C_p$ we have $\phi(\bar 1)^p = \phi(\bar 0) = \operatorname{id}_{C_q}$, so the only possible orders of $\phi(\bar 1)$ are $1, p$. On the other hand, every element in $\operatorname{Aut}(C_q)$---in particular $\phi(\bar 1)$ has order dividing $q - 1$, so if $\phi$ is nontrivial, $\phi(\bar 1)$ is not the identity automorphism, and thus $p \mid (q - 1)$. – Travis Willse Oct 31 '19 at 19:33