# Abstract Nonsense

## Review of Group Theory: Interesting Consequence of the First Isomorphism Theorem

Point of post: In this post we give one interesting “corollary” (it isn’t actually a corollary, but the tools used to prove it are all corollaries) of the First Isomorphism Theorem. Moreover, since it fits well with the post we’ll prove that if $G$ is a group w

Motivation

While the First Isomorphism Theorem may at first seem not that surprising, we shall see that it is used constantly to prove that two groups are isomorphic. Moreover, we shall see that it will be an integral part of proving the other isomorphism theorems. That said, there are some more ‘trivial’ corollaries of this theorem. We present a series of corollaries in succession to prove the neat theorem that if $N\trianglelefteq G$ and $\left(|N|,\left(G:N\right)\right)=1$ then $N$ is the unique subgroup of $G$ with order $|N|$. I picked this particular theorem since it uses a lot of the machinery we’ve covered as of yet.

Theorem: Let $G$ and $G'$ be groups and $G$ finite. Then,

$\left|G\right|=\left|\ker\phi\right|\left|\phi\left(G\right)\right|$

Proof: This follows immediately from the First Isomorphism Theorem and Lagrange’s theorem. More explicitly,

$\displaystyle \left|\phi(G)\right|=\left|G/\ker\phi\right|=\left(G:\ker\phi\right)=\frac{\left|G\right|}{\left|\ker\phi\right|}$. $\blacksquare$

From this we get the very interesting corollary:

Corollary: Let $G$ and $G'$ be finite groups with $\left(|G|,|G'|\right)=1$ then $\text{Hom}\left(G,G'\right)$ consists entirely if the trivial homomorphism $1:G\to G':g\mapsto e_{G'}$.

Proof: Note from our previous corollary that if $\phi\in\text{Hom}\left(G,G'\right)$

$|G|=\left|\ker\phi\right|\left|\phi\left(G\right)\right|$

and so in particular $\left|\phi\left(G\right)\right|{\large \mid} \left|G\right|$. But, by Lagrange’s theorem (since $\phi\left(G\right)\leqslant G'$) we have that $\left|\phi(G)\right|\mid |G'|$. Thus, $\left|\phi\left(G\right)\right|$ is a common divisor of $|G|$ and $|G'|$ and thus by assumption $\left|\phi\left(G\right)\right|=1$ from where the conclusion follows. $\blacksquare$

From this we get the following theorem

Theorem: Let $G$ be a finite group with $H\leqslant G$ and $N\trianglelefteq G$. If $\left(|H|,\left(G:N\right)\right)=1$ then $H\leqslant N$.

Proof: Note that since $N\trianglelefteq G$ we have that $N$ induces the canonical projection $\pi:G\to G/N$. Note though then that

$\pi_{\mid H}:H\to G/N$

is also a homomorphism. But, since $\left(|H|:\left|G/N\right|\right)=1$ we have from our previous corollary that $\pi_{H}(h)=N$ for every $h\in H$. But, this is true if and only if $h\in N$ for every $h\in H$ from where the conclusion follows. $\blacksquare$

Thus, we are finally able to prove our ‘neat’ theorem.

Theorem: Let $G$ be a finite group and $N\trianglelefteq G$ with $\left(|N|,\left(G:N\right)\right)=1$. Then, $N$ is the unique subgroup of $G$ with order $|N|$.

Proof: Suppose that $H\leqslant G$ with $|H|=|N|$. Then, $\left(|H|,\left(G:N\right)\right)=1$ and so $H\leqslant N$. But, since $|H|=|N|$ it follows that $H=N$. $\blacksquare$

And, as stated in the point of post, we prove a sort-of-but-not-really converse. Namely:

Theorem: Let $G$ be a group and $N\leqslant G$. If $N$ is the only subgroup of $G$ with order $|N|$ then $N\trianglelefteq G$.

Proof: We merely recall that for any $g\in G$ the inner automorphism $i_g$ is an automorphism and so in particular $i_g\left(N\right)\leqslant G$ and $\left|i_g\left(N\right)\right|=|N|$. It follows from our assumption that $i_g\left(N\right)=N$. $\blacksquare$

References:

1.  Lang, Serge. Undergraduate Algebra. 3rd. ed. Springer, 2010. Print.

2. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 200