I've been working on solving the following problem:
Is there an injective homomorphism between $G:=\mathbb{Z_4} \times \mathbb{Z_4}$ and
$S_7$?
I am at a loss as to how to solve this.
I've considered a few things:
- (trying to refute the claim): If there were a homomorphism $\phi$ between $\mathbb{Z_4} \times \mathbb{Z_4}$ and $S_7$, then for all $g \in \mathbb{Z_4} \times \mathbb{Z_4}$ we would have $o(\phi(g)) \mid o(G)=16$. In particular, there must be a permutation $a\in S_7$ such that $o(a) \mid 16$. Indeed there is – e.g. $(1 2)(34)(56)(7)$ is of order $2$ (that's the ${\rm lcm}$).
- Using Cayley's theorem somehow (e.g. by showing that there's a subgroup $H\leq G$ such that $[G:H]=7$ and then by Cayley we get a homomorphism, but: a. I don't see how this can be true and b. still not injective).
- Something to do with the kernel since injective means the kernel is trivial: $K=\{(0,0)\}$, but I don't immediately see why this is helpful.
Any advice? What am I missing?
Thanks.
Best Answer
Here’s a more conceptual solution which I think is cleaner. How many times is $7!$ divisible by $2$? A quick check yields $2^4$, so a $2$-Sylow subgroup of $S_7$ has order $16$.
Remember that all $2$-Sylow subgroups are conjugate, hence isomorphic. Your task, then, is to find such a group, and show that it is not isomorphic to $C_4\times C_4$, the product of two cyclic groups.
Constructing a group of order $16$ inside $S_7$ isn’t very hard: you should be familiar by now with dihedral groups. A dihedral group of size $8$ is easy to spot inside $S_7$, and then just multiply it by a cyclic group of order $2$ to obtain a Sylow subgroup. Finally, observe that this group isn’t isomorphic to $C_4\times C_4$ by counting elements of various orders (or, if you’re less silly than I am, by observing that one of these groups is abelian and the other is not.)