How do I know that $\mathbb{Z}_{175}$ is not an additional subgroup of order $175$?

Question

Here was the original problem statement.

Enumerate all non-isomorphic groups of order $175$.

See that $|G| = 175 = 5^2\cdot7$. Therefore by Sylow's first $H \leq G$ & $|H| = 25$ in addition to $K \leq G$ & $|K| = 7$. By Sylow's third $n_5 \equiv 1\pmod{5}$ and $n | 7$, so $n_5 = 1$ and $H$ is normal. By Sylow's third $n_7 \equiv 1\pmod{7}$ and $n| 25$, so $n_7 = 1$ and $H$ is normal. So both $H$ and $K$ are normal subgroups. We also know that $H \cap K = \{e\}$ because $\gcd(5, 7) = 1$.

Because $H \lhd G$ or $K \lhd G \Rightarrow HK \leq G$, therefore $HK \leq G$. $|HK| = 25 \cdot 7$ so $HK \cong G$. Because both groups are normal, $G \cong HK \cong H \times K$. Because $|H| = p^2 = 5^2$, therefore $G \cong \mathbb{Z}_5 \times \mathbb{Z}_5 \times \mathbb{Z}_7$ or $G \cong \mathbb{Z}_{25} \times \mathbb{Z}_7$.

But my proof is incomplete. I said that $G$ must be isomorphic to either $\mathbb{Z}_5 \times \mathbb{Z}_5 \times \mathbb{Z}_7$ or $\mathbb{Z}_{25} \times \mathbb{Z}_7$... but it could be possible that all groups of order 175 are isomorphic to only $\mathbb{Z}_{25} \times \mathbb{Z}_7$ and never $\mathbb{Z}_5 \times \mathbb{Z}_5 \times \mathbb{Z}_7$ . So I haven't shown that the group $\mathbb{Z}_5 \times \mathbb{Z}_5 \times \mathbb{Z}_7$ isn't an extraneous solution! How do I show that these groups are non-distinct?

Answer 1

. . . it could be possible that all groups of order $175$ are isomorphic to only $\Bbb{Z}_{25} \times \Bbb{Z}_7$ and never $\Bbb{Z}_5 \times \Bbb{Z}_5 \times \Bbb{Z}_7$. So I haven't shown that the group $\Bbb{Z}_5 \times \Bbb{Z}_5 \times \Bbb{Z}_7$ isn't an extraneous solution! How do I show that these groups are non-distinct?

There are really two questions here:

1. Is $\Bbb{Z}_{25} \times \Bbb{Z}_7$ in fact isomorphic to $\Bbb{Z}_5 \times \Bbb{Z}_5 \times \Bbb{Z}_7$?
2. If not, can I find groups $G$ and $H$ such that $G \cong \Bbb{Z}_{25} \times \Bbb{Z}_7$ and $H \cong \Bbb{Z}_5 \times \Bbb{Z}_5 \times \Bbb{Z}_7$?

The answer to the first question is no, they are not isomorphic. As @Mark points out in the comments below his answer, $\Bbb{Z}_{25} \times \Bbb{Z}_7$ has an element of order $25$, namely $(1,0)$, whereas there is no element of order $25$ in $\Bbb{Z}_5 \times \Bbb{Z}_5 \times \Bbb{Z}_7$. Hence, the two groups are non-isomorphic.

The answer to the second question is, yes: simply take $G$ to be $\Bbb{Z}_{25} \times \Bbb{Z}_7$ and $H$ to be $\Bbb{Z}_5 \times \Bbb{Z}_5 \times \Bbb{Z}_7$. That is, since every group is isomorphic to itself, we know that there is at least one group isomorphic to $\Bbb{Z}_{25} \times \Bbb{Z}_7$ (namely, itself) and there is at least one group isomorphic to $\Bbb{Z}_5 \times \Bbb{Z}_5 \times \Bbb{Z}_7$, (namely, itself).

---

So, the map $\{ G : |G| = 175 \} \to \{ \Bbb{Z}_{25} \times \Bbb{Z}_7, \Bbb{Z}_5 \times \Bbb{Z}_5 \times \Bbb{Z}_7 \}$ which sends $G$ to its isomorphic copy in the codomain is a well-defined map into a two-element set that is also surjective:

• In your proof, you have shown that every group of order $175$ is isomorphic to either $\Bbb{Z}_{25} \times \Bbb{Z}_7$ or $\Bbb{Z}_5 \times \Bbb{Z}_5 \times \Bbb{Z}_7$. (Well-definedness)
• In the answer to question 1. you have shown that these are in fact distinct groups of order $175$. (Codomain is a two-element set)
• In the answer to question 2. you have shown that there is at least one group that is isomorphic to each of these groups. (Surjectivity)

Answer 2

You took a group of order $175$ (without making any other assumptions about the group) and you proved that it must be isomorphic to one of these two groups. So I don't really understand the question. $\mathbb{Z_{175}}$ is isomorphic to $\mathbb{Z_{25}}\times\mathbb{Z_7}$, it follows from the Chinese remainder theorem.

Answer 3

The group $H$ acts on $K = \mathbb{Z}_7$ by conjugation. Since $\#H = 25$ is prime to the order of $\operatorname{Aut}(\mathbb{Z}_7) = \mathbb{Z}_6$, that action must be trivial; that is, $H$ and $K$ commute, and $G = H \times K$. For $p$ prime, the only group of order $p$ is $\mathbb{Z}_p$, and the only groups of order $p^2$ are $\mathbb{Z}_p^2$ and $\mathbb{Z}_p \oplus \mathbb{Z}_p$. (For the latter, note that any group of order $p^2$ has nontrivial center, and the only action of $\mathbb{Z}_p$ on itself is trivial by the same argument as for $K$ above.)