# For all subgroups $K \leq U(65)$, $K \ncong \mathbb Z_2 \oplus \mathbb Z_2 \oplus \mathbb Z_2$.

How would I show that for all subgroups $$K \leq U(65)$$, $$K \ncong \mathbb Z_2 \oplus \mathbb Z_2 \oplus \mathbb Z_2$$?

I know that $$U(65) \cong \mathbb Z_4 \oplus \mathbb Z_{12}$$, in which $$\mathbb Z_4 \oplus \mathbb Z_{12}$$ has three elements of order 2 and $$\mathbb Z_2 \oplus \mathbb Z_2 \oplus \mathbb Z_2$$ has seven elements of order 2. How does that deduce that $$K \ncong \mathbb Z_2 \oplus \mathbb Z_2 \oplus \mathbb Z_2$$ for all subgroups $$K$$ of $$U(65)$$?

This is just what I'm reading from an answer text, which doesn't make much sense.

• Note: $U(65)$ is the multiplicative group of units modulo $65$ Commented Oct 30, 2019 at 12:41
• I know, but I don't know what this would imply.
– Tim
Commented Oct 30, 2019 at 12:43
• That note is for the others reading the question. I was personally not used to the notation, so I had to take an educated guess and google to verify. Commented Oct 30, 2019 at 12:44
• To tack on to Omnomnomnom's answer (what a mouth full) if it wasn't already obvious, group isomorphisms must preserve group structure, including the order of the elements. Commented Oct 30, 2019 at 12:45

The point being made is this: consider a fixed subgroup $$K \subset G$$. If $$K \cong \Bbb Z_2 \oplus \Bbb Z_2 \oplus \Bbb Z_2$$, then $$K$$ must also have seven elements of order $$2$$. However, since $$K$$ is a subset of $$U(65)$$, $$K$$ can have at most $$3$$ elements of order $$2$$.
Exercise. If $$A$$ and $$B$$ are isomorphic groups then the number of elements of $$A$$ of order two is the same as the number of elements of $$B$$ of order two.
Important Note: If the exercise is not obvious you haven't quite got what "isomorphic" means. Anything you say about $$A$$ ("in the language of group theory") that's true of $$A$$ is also true of $$B$$, because they are in some sense the same group! Maybe $$B$$ consists of the elements of $$A$$ painted black...
Hint: Say $$f:A\to B$$ is an isomorphism. Show that $$x\in A$$ has order two if and only if $$f(x)\in B$$ has order two.