## 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 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 and then is the unique subgroup of with order . I picked this particular theorem since it uses a lot of the machinery we’ve covered as of yet.

**Theorem: ***Let and be groups and finite. Then,*

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

.

From this we get the very interesting corollary:

**Corollary: ***Let and be finite groups with then consists entirely if the trivial homomorphism .*

**Proof: **Note from our previous corollary that if

and so in particular . But, by Lagrange’s theorem (since ) we have that . Thus, is a common divisor of and and thus by assumption from where the conclusion follows.

From this we get the following theorem

**Theorem: ***Let be a finite group with and . If then .*

**Proof: **Note that since we have that induces the canonical projection . Note though then that

is also a homomorphism. But, since we have from our previous corollary that for every . But, this is true if and only if for every from where the conclusion follows.

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

**Theorem: ***Let be a finite group and with . Then, is the unique subgroup of with order .*

**Proof: **Suppose that with . Then, and so . But, since it follows that .

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

**Theorem: ***Let be a group and . If is the only subgroup of with order then .*

**Proof: **We merely recall that for any the inner automorphism is an automorphism and so in particular and . It follows from our assumption that .

**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

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