There appears in Seth Warner's "Modern Algebra" Exercises 12.15 what seems to be an obvious truth:

If $H$ and $K$ are normal subgroups of a group $G$ such that $H \subseteq K$, then $H$ is a normal subgroup of $K$, ...

Indeed it is so:

$\forall g \in G: g K = K g$ by definition of a normal subgroup

$\forall g \in G: g H = H g$ by definition of a normal subgroup

$ \leadsto \forall g \in K: g H = H g$ as $K \subseteq G$

That seems too easy.

In fact, it even appears that if $H$ is a subset of any subgroup $K$ -- or even if $K$ is merely a subset -- of $G$, then $H$ is a normal subgroup of $K$.

This result is rarely seen in the elementary literature on group theory and abstract algebra, or at least, in any of the works I've seen. But so obvious is it, and so useful is it, that I wonder whether I'm made a silly mistake in understanding this, and in fact it is not such a trite and trivial result after all.

Can my thinking be confirmed -- that in fact a normal subgroup $H$ of a group $G$ is also a normal subgroup of any subset $K$ such that $H \subseteq K \subseteq G$? And if not, where am I going wrong in my thinking?

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – TheSimpliFire
    Commented Mar 19, 2022 at 7:38

1 Answer 1


Your argument is correct.

The result appears in many books. For example:

  • J.J. Rotman, Introduction to the Theory of Groups, 4th Edition. The 9th Exercise in the section on normal subgroups (the section only contains the definition), page 31, Exercise 3.33: "If $K\leq H\leq G$ and $K\triangleleft G$, then $K\triangleleft H$."

  • W.R. Scott's Group Theory, from 1964. Page 29, Exercise 2.1.24(c): "If $H\triangleleft G$ and $H\subset K\subset G$, then $H\triangleleft G$." (Scott uses $\subset$ to denote subgroups that need not be proper, reserving $\lt$ for proper subgroups).

Likely you have not spotted it because the more general version that is most useful is the following:

Let $G$ be a group, and $N\triangleleft G$ a normal subgroup of $G$. Then for any subgroup $H$ of $G$, $N\cap H\triangleleft H$.

In fact, this statement is part of the Second Isomorphism Theorem (which states that $NH/N\cong H/(N\cap H)$). In the special case where $N\leq H\leq G$, you get your result: $N\cap H=N\triangleleft H$.

In this form, it appears in:

  • Dummit and Foote, Abstract Algebra Second Edition. Exercises on Chapter 3, Quotient Groups and Homomorphisms, has Exercise 24, p. 89.

  • Lang's Algebra, Revised 3rd Edition, mentions en passant on page 17: "Let $G$ be a group, and let $H,K$ be two subgroups. Assume that $H$ is contained in the normalizer of $K$. Then $H\cap K$ is obviously a normal subgroup of $H$[.]" This yields this result when $K$ is normal and $H$ contains $K$.

  • Derek Robinson's A Course in the Theory of Groups 2nd Edition, 1.4.4 (on page 19), the Second Isomorphism Theorem: "Let $H$ be a subgroup and $N$ a normal subgroup of a group $G$. Then $N\cap H\triangleleft H$ and $(N\cap H)x\longmapsto Nx$ is an isomorphism from $H/(N\cap H)$ to $NH/N$."

  • Jacobson's Basic Algebra I, Theorem 1.9 (page 64): "Let $H$ and $K$ be subgroups of $G$, $K$ normal in $G$. Then $HK=\{hk\mid h\in H, k\in K\}$ is a subgroup of $G$ containing $K$, $H\cap K$ is normal in $H$, and the map $hK\mapsto h(K\cap H)$, $h\in H$, is an isomorphism of $HK/K$ with $H/(H\cap K)$.

  • van der Waerden's Algebra, volume I, Section 7.3, page 147; he uses "divisor" rather than subgroup: "If $N$ is a normal divisor in $G$, $K$ is a subgroup $G$, then the intersection $N\cap K$ is a normal divisor in $K$[.]"

etc. Since this fact is key to one of the Isomorphism Theorems, I dare say that any book on Group Theory that includes the Isomorphism Theorems will necessarily include it, your result being a special case.

  • $\begingroup$ Why is the S.I.T. "more general"? I'm missing something. $\endgroup$ Commented Mar 18, 2022 at 15:43
  • 1
    $\begingroup$ @PrimeMover: The statement "If $N\triangleleft G$ and $K$ is any subgroup of $G$, then $N\cap K\triangleleft K$" is more general than "If $N\triangleleft G$ and $K$ is a subgroup of $G$ with $N\leq K\leq G$, then $N\triangleleft K$." Not the Second Isomorphism Theorem,, but just the statement about $N\cap K$ being normal in $K$. $\endgroup$ Commented Mar 18, 2022 at 15:49

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