Number of non isomorphic groups of order $122$, My attempt through Sylow theory.

Question

$|G|=122 = 2 . 61$

No. of sylow $2$ subgroups $= 1$ or $61 = n_2$

No. of sylow $61$ subgroups $= 1 = n_{61}$

Let the group of order $61$ be $H_{61}$ and the group of order $2$ be $H_2$

Then : clearly both $H_{61}$ and $H_2$ are cyclic.

and if $a ∈ H_2 ⋂ H_{61}$

then, since $|a|$ must divide simultaneously both $2$ and $61$

hence, $H_{61} ⋂ H_2 = \{{e}\}$

Since, $n_{61}=1$ hence, $H_{61}$ is normal and $( H_{61} H_2 )$ is a subgroup $.....(1)$

Now, $| H_{61} H_{2} | = 61 . 2 = 122 ..... (2)$

From $(1)$ and $(2)$ :$ H_{61} H_2 = G ........ (3)$

Since $G = H_2 H_{61}$ and $|G|=122 = |H_2||H_{61}|$, there cannot be any more subgroups of order $2$. Hence $n_2$ must be equal to $1$

Hence, from $(1), (2), (3) : G = H \times K ≅ H ⊕ K ≅ \mathbb Z_2 ⊕ Z_{61} ≅ Z_{122}$

What could have gone wrong in my method? Why am I not getting $D_{61}$ as an answer too?

EDIT : A theorem from Gallian says this :

If a group $G$ is the internal direct product of a finite nymber of subgroups $H_1,H_2, \cdots H_n$, then

$G \approx H_1 \oplus H_2 \oplus \cdots \oplus H_n$

Hence, here $H_2 \approx Z_2$ and $H_{61} \approx Z_{61}$ .

Hence, $G = H \times K \approx H \oplus K \approx Z_2 \oplus Z_{61} \approx Z_{122}$

Where could I have gone wrong?

Thank you for your help.

Answer 1

You are correct that $G = H_2 H_{61}$, but this is not a direct product unless $H_2$ is normal, or equivalently, $n_2 = 1$.

If $G = H_2 H_{61}$ is a direct product, then its elements are (under the appropriate isomorphism) ordered pairs of the form $(h,k)$ where $h \in H_2$ and $k \in H_{61}$. If we have $(h,k)(h,k) = (1,1)$ for some element $(h,k)$, then by definition of multiplication in a direct product, this means $(h^2, k^2) = (1,1)$, which is true if and only if $h^2 = 1$ and $k^2 = 1$. Since $k \in H_{61}$ this forces $k=1$, so $(h,k) = (h,1)$. In other words, the only element of $H_2 H_{61}$ with order $2$ is the nonidentity element of $H_2$, or putting it another way, $H_2$ is the unique (hence normal) subgroup of order 2.

But note that $D_{61}$ is also expressible in the form $H_2 H_{61}$. The difference here is that $H_2$ is not normal. Indeed, if $f$ denotes a flip about one of the axes of symmetry, and $r$ is a rotation by one position (say clockwise), then $\langle r^k f\rangle$ is a subgroup of order 2 for every $k$, so $n_{2} = 61$.

It remains to check that $D_{61}$ is the only group (up to isomorphism) of the form $H_2 H_{61}$ where $H_{61}$ is normal and $H_2$ is not.

To do this, one option is to note that if $n_2 = 61$, then I can certainly choose two distinct elements of order $2$. If I can show that $G$ is generated by these two elements, then I can conclude that $G$ must be dihedral, because any finite group generated by two involutions is dihedral. However, if you haven't already proved that result, then this just kicks the can down the road.

Another option is to try to prove directly that $G$ must satisfy the defining relations for $D_{61}$: namely, it is generated by two (not necessarily unique) elements $r$ and $s$ which satisfy $r^{61} = 1$, $s^2 = 1$, and $s^{-1}rs = r^{-1}$.

Answer 2

I hope you don't mind if I add a second answer. This is something of a side note and I don't want to clutter the main answer with it, but I think it might help clarify an issue you raised in the comments.

If $G = H_2 H_{61}$, and $H_2$ is not normal, you asked where the other elements of order $2$ are hiding. Let's explore this a bit.

First, note that $H_2$ is a subgroup of order $2$, and $H_{61}$ is a subgroup of order $61$. Their intersection is the identity. So, between them, they account for $62$ elements out of $122$: in other words, $|H_2 \cup H_{61}| = 62$. (Note that $H_2 \cup H_{61}$ is just a set, not a group.)

Now consider the set of all the remaining elements: $S = G \setminus (H_2 \cup H_{61})$. What are the orders of the elements in $S$? There two cases: either $H_2$ is normal or it is not. ($H_{61}$ is always normal since it has index 2.)

Case 1: $H_2 \lhd G$. In this case, $H_2$ contains the only element of order $2$, and $H_{61}$ contains the only elements of order $61$. Additionally, the identity is not in $S$, so by Lagrange's theorem, all the elements in $S$ must have order $122$. We conclude that $G$ is cyclic, and an element generates $G$ if and only if it is in $S$.

Case 2: $H_2$ is not normal. In this case, $G$ cannot be cyclic, because cyclic $\implies$ abelian $\implies$ all subgroups are normal. Therefore there are no elements of order $122$. As before, $H_{61}$ contains the only elements of order $61$, and the identity is not in $S$. Therefore all elements in $S$ must have order $2$. Since $|S| = 122 - 62 = 60$, this means we have found $60$ additional subgroups of order $2$, which along with the original $H_2$ gives us $61$ total.